FRANKLIN  INSTITUTE  LIBRARY 


obtained  the  sanction  of  the  Committee.  The  second  class  shall  include 
those  books  intended  for  circulation. 

Article  \ I. — The  Secretary  shall  have  authority  to  loan  to  Members 
and  to  holders  of  second  class  stock,  any  work  belonging  to  the  second 
class,  subject  to  the  following  regulations  : 

Section  1. — No  individual  shall  be  permitted  to  have  more  than  two 
7J ooks  out  at  one  time,  without  a written  permission,  signed  by  at  least 
two  members  of  the  Library  Committe ; nor  shall  a book  be  kept  out 
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rower may  renew  the  loan.  Should  any  person  have  applied  for  it,  the 
latter  shall  have  the  preference. 

Section  2. — A fine  of  ten  cents  per  week  shall  be  exacted  for  the 
detention  of  a book  beyond  the  limited  time  ; and  if  a book  be  not  re- 
turned within  three  months  it  shall  be  deemed  lost,  and  the  borrower 
shall,  in  addition  to  his  fines,  forfeit  its  value. 

Section  3.-—  Should  any  book  be  returned  injured,  the  borrower  shall 
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borrower  shall  replace  them  or  make  full  restitution. 

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sion from  the  proper  authorities,  any  book,  newspaper  or  other  property 
in  charge  of  the  Library  Committee,  shall  be  reported  to  the  Committee, 
who  may  inflict  any  fine  not  exceeding  twenty-five  dollars. 

Article  A ill. — No  member  or  holder  of  second  class  stock,  whose 
• annual  contribution  for  the  current  year  shall  be  unpaid  or  who  is  in 
arrears  for  fines,  shall  be  entitled  to  the  privileges  of  the  Library  or 
Reading  Room. 

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refuse  or  neglect  to  comply  with  the  foregoing  rules,  it  shall  be  the  duty 
of  the  Secretary  to  report  him  to  the  Committee  on  the  Library. 

Article  X.- — -Any  Member  or  holder  of  second  class  stock,  detected 
in  mutilating  the  newspapers,  pamphlets  or  books  belonging  to  the  Insti- 
tute shall  be  deprived  of  his  right  of  membership,  and  the  name  of  the 
offender  shall  be  made  public. 


PHILADELPHIA 


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Digitized  by  the  Internet  Archive 
in  2015 


https://archive.org/details/carpentrymadeeas00bell_0 


OR,  THE 


SCIENCE  AND  ART  OF  FRAMING, 

OK  A 

NEW  AND  IMPROVED  SYSTEM. 


WITH  SPECIFIC  INSTRUCTIONS  FOR 

BUILDING  BALLOON  FRAMES,  BARN  FRAMES,  MILL  FRAMES,  WARE- 
HOUSES, CHURCH  SPIRES,  ETC. 


COMPRISING  ALSO 


4’  SYSTEM  OF  BRIDGE  BUILDING; 

* 1 ' » ’ •*  * « ■ j \ * » V ->  ' > * 


Bills,  Estiiaates  oe\  Oqs,iv^:tst:q  Valuable  Tables. 


Unty-four  Plates  and  neat  Sun  fhtndttfl  Jigutti 


SECOND  EDITION,  ENLARGED  AND  IMPROVED. 


BY  WILLIAM  E.  BELL, 

ARCHITECT  AND  PRACTICAL  BUILDER. 


PHILADELPHIA : 

PEBausoisr  bros.  s&  co. 

1889. 


\ 


Entered  according  to  Act  of  Congress,  in  the  year  1857,  by 
WILLIAM  E.  BELL, 

In  the  Clerk's  Office  of  the  District  Court  of  the  United  States  in  and  for  the  Eastern  District  of  Pennsylvania. 


Entered  according  to  Act  of  Congress,  in  the  year  1875,  by 
WILLIAM  E.  BELL, 

"in  the"  Offict  ofithu  Librarian  of  Congress,  at  Washington  ,'D.  C, 


FERGUSON  BROS.  & CO., 
PRINTERS  AND  ELECTROTYPERS, 
PHILADELPHIA. 


PREFACE  TO  THE  SECOND  EDITION. 


The  “ Art  and  Science  of  Carpentry  ” was  first  presented  to 
the  public  in  1858.  I then  made  some  efforts  to  introduce  it 
into  general  use,  and  especially  to  bring  it  to  the  notice  of 
mechanics  and  their  apprentices.  At  first  these  efforts  met 
with  only  partial  success.  Then  the  cry  of  “ hard  times  ” was 
no  humbug;  many  good  workmen  were  out  of  employment 
and  money  was  really  scarce.  The  public  has  also  been  so 
often  imposed  upon  by  charlatans  and  quacks  of  all  kinds  that 
it  is  always  difficult  to  introduce  anything,  however  valuable, 
until  the  popular  mind  has  had  time  to  be  convinced  of  its  real 
utility.  It  was  not  long  before  orders  for  the  work  began  to 
come  in,  and  the  first  issue  of  a thousand  copies  was  exhausted. 
Another  thousand  was  then  issued  and  sold,  and  the  third  was 
commenced  when  the  publishers  and  my  friends  generally,  who 
had  become  fully  convinced  of  the  value  of  the  work,  urged  me 
to  prepare  a revised  and  improved  edition.  This  I have  now 
undertaken,  and  the  work  is  here  presented,  revised,  enlarged, 
and  improved  in  all  its  departments,  in  hopes  of  rendering  it 
more  intrinsically  valuable,  as  well  as  more  acceptable  and 
useful. 

Young  men  are  apt  to  regard  the  carpenter’s  trade  an  easy 
one,  but  there  never  was  a greater  mistake,  for  I believe  it  from 
experience  to  be  one  of  the  hardest.  When  I had  finished  my 
apprenticeship  my  old  master  assured  me  that  I had  now  fully 
learned  my  trade  and  was  myself  a master-workman,  and  1 
was  elated  with  the  thought  that  I knew  it  all ; but  on  travel- 
ling to  other  places  and  working  in  other  shops  I was  soon 

(iii) 


S 57<2| 


IV 


PREFACE. 


made  to  feel  my  deficiencies,  and  from  that  time  to  this  I have 
applied  my  mind  to  the  study  of  my  business,  with  what 
success  I leave  the  public  to  judge. 

In  this  second  edition  I have  done  my  utmost  to  make  the 
work  so  plain  that  a common  school-boy  can  understand  the 
rules  and  appreciate  their  value.  In  contrast  with  this  sim- 
plicity, I have  now  before  me  a book,  just  published,  which  goes 
back  to  the  old  way  of  obtaining  the  lengths  and  bevels  of  dif- 
ferent parts  of  a frame  by  lines  and  tangents  which  I am  cer- 
tain that  not  one  carpenter  in  twenty  can  use  understandingly, 
and  this  confirms  me  in  the  importance  of  the  remarks  on  this 
subject  contained  in  the  introductory  chapter.  Only  a few 
days  ago,  a master  builder  remarked  to  me  that  the  rule  here 
given  for  finding  the  backing  on  hip  rafters  was  alone  worth 
the  price  of  the  book,  and  this  is  only  one  among  many  im- 
portant rules  herein  contained,  some  of  which  I judge  to  be  of 
greater  value  than  this. 

I have  frequently  been  asked  why  there  is  so  much  geometry 
introduced  into  this  book  ? For  an  answer  to  this  question,  I 
beg  every  one  to  read  the  introductory  chapter.  The  rules  of 
carpentry  are  founded  upon  geometry,  and  it  is  only  by  a course 
of  geometrical  reasoning  that  these  rules  can  be  demonstrated, 
and  only  as  much  of  geometry  is  here  given  as  is  necessary  for 
this  purpose. 

The  work  having  been  stereotyped  on  its  first  issue,  we  now 
insert  the  new  matter  at  the  end,  which  will  add  very  materially 
to  its  value.  We  have  also  revised  the  whole  work  carefully, 
and  corrected  all  errors  of  any  importance,  and  it  is  here  pre- 
sented with  the  author’s  thanks  for  past  encouragement  and 
appreciation,  and  his  hopes  for  their  renewal  and  continuation. 

The  Author. 


Ottawa,  Illinois. 


CONTENTS 


PART  I. — Geometry. 

Definitions 17 

Explanations  of  Mathematical  Symbols 21 

Definitions  of  Mathematical  Terms 22 

Axioms 22 

Proposition  I.  Theorem 23 

Proposition  XXX.  Problem 38 

PART  II — Carpentry 

Use  op  the  Square  in  Obtaining  Bevels 4 3 

The  square  described 43 

Pitch  of  the  roof 43 

Bevels  of  Rafters 44 

Bevels  of  upper  joints  and  gable-end  studding 45 

Bevels  of  Braces 45 

Balloon  Frames 47 

The  sills  (light  sills) 47 

The  studs 47 

The  plates 4 7 

Raising  and  plumbing  the  frame 47 

The  floor  joists 48 

Upper  joists 48 

Rafters 48 

Gable-end  studs 49 

Framing  the  sills 50 

Work  sides  (of  timbers) 50 

To  take  timber  out  of  wind 50 

Spacing  for  windows  and  doors ...... 51 

Mortices  for  the  studs 51 

The  gains  (for  joints) 51 

The  draw  bores. 51 

A draw  pin 51 

Supports  for  the  upper  joists 52 

Crowning  of  joists 53 

Bridging  of  joists 53 

Lining,  or  sheeting  balloon  frames 53 

Bi  rn  Frames 55 

Size  of  mortices 55 

Braces 56 

Pitch  of  the  roof. 56 

Purlins . 56 

Length  of  the  purlin  posts 57 

Purlin  post  brace 58 

Purlin  post  brace  mortices 58 

Upper  end  bevel  of  purlin  post  braces 60 

Mill  Frames 61 

Cripple  studs 61 

Trussed  parti  Cons 6> 


(5) 


6 CONTENTS. 

nai 

Scarfing 64 

Straps  and  bolts  (in  scarfing)..., 64 

Scarfing  over  posts 65 

Floors  in  Brick  Buildings 66 

Trimmer  joists. , 66 

Circular  Centres 67 

Elliptical  Centres 68 

Arches 69 

Hip  Roofs 76 

Hip  Rafters 70 

Side  bevel  of  hip  rafters 70 

Down  bevel  of  hip  rafters 71 

Backing  of  hip  rafters 71 

Lengths  and  bevels  of  the  jack  rafters ..  72 

Hips  and  Valleys 73 

Trapezoidal  Hip  Roofs 73 

Lengths  of  the  irregular  hip  rafters 73 

Bevels  of  the  irregular  hip  rafters 74 

Backing  of  hip  rafters  on  trapezoidal  and  other  irregular  roofs 75 

Length  of  jack  rafters 75 

Side  bevels  of  jack  rafters  on  the  sides  of  the  frame 76 

Side  bevels  of  the  jack  rafters  on  the  slant  end  of  the  frame 76 

Down  bevel  of  the  jack  rafters  on  the  beveled  end  of  the  frame 77 

Octagonal  and  Hexagonal  Roofs 78 

Length  of  the  hip  rafters 78 

Bevels  of  the  hip  rafters 78 

Backing  of  the  octagonal  hip  rafters 79 

Length  of  the  jack  rafters 79 

Width  of  the  Building 79 

Tvoofs  of  Brick  and  Stone  Buii, dings 81 

Lengths  and  bevels  of  the  braces 81 

Dimensions  of  timbers  for  figs.  1 and  2 82 

Length  of  straining  beam 83 

Church  Spires 89 

Domes 99 

PART  III.— Bridge  Building. 

Straining-Beam  Bridges . 93 

Trestle  Bridges 98 

Arch  Truss  Bridges * 102 

General  Principles  of  Bridge  Building 107 

Backing  of  hip  rafters  demonstrated Ill 

Bracing  between  posts  that  stand  battering.... 113 

Bevels  of  the  braces 114 

Height  of  the  girder 114 

Length  of  the  girder  and  cap 114 

The  long  braces 114 

How  to  calculate  the  length  of  the  braces 115 

Improved  straining-beam  bridge 116 

Bill  of  timber 116 

Bill  of  iron .' 117 

Large  circular  center  for  heavy  arches 1 1 8 

Husk  or  trestle-work  for  water-tank 119 

Height  and  location  of  tank ...  119 

To  estimate  the  capacity  of  a tank 1 19 

Framing  and  constructing  pile  bridges 122 

PART  IV. — Explanation  of  the  Tables. 

Definitions  of  Terms  and  Phrases  used  in  this  Work 127 

Table  I.  Length  of  Common  Rafters 120 

Table  II.  Length  of  Hip  Rafters I S3 

Table  III.  Octagonal  Hoofs . 129 

Table  IV  Length  of  Braces Ill 

Table  V.  Weight  of  Square  Iron I 12 

Table  VI  Weight  of  Flat  Iron 144 

Table  VII.  Weight  of  Round  Iron I 16 

Table  V 1 1 {.  Weight  and  Strength  of  Timber 148 

How  to  Calculate  the  Strength  of  Timber U9 


INTRODUCTORY  CHAPTER. 


SUMMARY  VIEW. 


o 


The  Science  and  the  Art  oi  IB  raining. 

No  apology  is  offered  for  introducing  to  the  Public  a work 
on  the  Science  and  Art  of  Framing.  By  the  Science  of  Fram- 
ing is  meant  the  certain  knowledge  of  it,  founded  on  mathe- 
matical principles,  and  for  which  the  master  of  it  can  assign 
intelligent  reasons,  which  he  knows  to  be  correct ; while  the 
Art  of  Framing  is  the  system  of  rules  serving  to  facilitate  the 
practice  of  it,  but  the  reasons  for  which  the  workman  may  or 
may  not  understand.  That  Carpentry  has  its  rules  of  Science 
as  well  as  its  rules  of  Art,  no  intelligent  mechanic  can  doubt. 
The  rules  of  the  Art  are  taught  by  the  master-workman  at 
the  bench  ; or,  more  commonly,  insensibly  acquired  by  habit 
and  imitation.  But  by  whom  have  the  rules  of  the  Science, 
been  laid  down,  and  where  have  its  principles  been  intelli- 
gibly demonstrated? 

Something  New. 

It  is  believed  that  this  is  the  very  first  attempt  ever  made 
to  bring  the  Science  of  Carpentry,  properly  so  called,  within 
the  scope  of  practical  mechanics. 


(7) 


3 


CARPENTRY  MADE  EASY. 


Deficiencies  of  Former  Works  on  Carpentry. 

Whatever  has  formerly  been  published  on  this  subject,  that 
can,  with  any  degree  of  propriety,  be  classed  under  the  head 
of  Science,  has  been  only  available  by  professional  Architects 
and  Designers,  being  written  in  technical  language  and  mathe- 
matical signs,  accompanied  by  no  adequate  definitions  or  ex- 
planations ; and  are  as  perfectly  unintelligible  to  working-men 
of  ordinary  education  as  Chinese  or  Choctaw.  On  the  other 
hand,  the  numerous  works  upon  the  Art  of  Carpentry,  de 
signed  and  published  for  the  use  of  working-men,  are  sadly 
deficient  in  details  and  practical  rules.  They  seem  to  take 
it  for  granted  that  the  student  is  already  familiar  with  his 
business;  they  furnish  him  with  drafts  and  plans  to  work 
from;  they  tell  him  authoritatively  that  such  or  such  an 
angle  is  the  proper  bevel  for  such  a part  of  the  frame ; but 
they  neither  tell  him  why  it  is  so,  nor  inform  him  how  to 
begin  and  go  on  systematically  with  framing  and  erecting  a 
building.  These  works  are,  in  fine,  chiefly  valuable  for  their 
plates ; and  even  these  it  is  not  always  possible  to  work  from 
with  confidence  and  accuracy,  because  no  man  can  work  with 
confidence  and  accuracy  in  the  dark : and  he  certainly  is  in 
the  dark  who  does  not  understand  the  reasons  on  which  his 
rules  are  founded. 


The  Author’s  Experience. 

These  facts  and  reflections  have  been  impressing  themselves 
upon  the  mmd  of  the  Author  of  this  work  for  twenty  years 
past,  while  he  has  been  serving  the  Public  as  a practical  car 
penter.  During  much  of  this  time  it  has  been  his  fortune 
to  have  larg°  jobs  on  hand,  employing  many  journeymen 


INTRODUCTORY  CHAPTER, 


9 


mechanics,  who  claimed  to  understand  their  trade,  and  de- 
manded full  wages.  But  it  has  been  one  of  the  most  serious 
and  oppressive  of  his  cares,  that  these  journeymen  knew  so 
little  of  their  business. 

Few  Good  Carpenters. 

They  had,  by  habit,  acquired  the  use  of  tools,  and  could 
perform  a job  of  work  after  it  had  been  laid  out  for  them; 
but  not  more  than  one  man  in  ten  could  himself  lay  out  a 
frame  readily  and  correctly. 

Why  Apprentices  do  not  Learn. 

Now,  it  is  not  commonly  because  apprentices  are  unwilling 
to  learn,  or  incapable  of  learning,  that  this  is  so,  but  it  is  be- 
cause they  have  not  the  adequate  instruction  to  enable  them 
to  become  master-workmen.  Their  masters  are  very  natu- 
rally desirous  to  appropriate  their  services  to  their  own  best 
advantage ; and  that  is  often  apparently  gained  by  keeping 
the  apprentice  constantly  at  one  branch  of  his  business,  in 
which  he  soon  becomes  a good  hand,  and  is  taught  but  little 
else ; and  when  his  time  is  his  own,  and  he  comes  to  set  up 
business  for  himself,  then  he  is  made  to  feel  his  deficiencies. 
Should  he  have  assistants  and  apprentices  in  his  turn,  he 
would  be  unable  to  give  them  proper  instruction,  even  were 
he  well  disposed  to  do  so— for  he  can  teach  them  nothing 
more  than  what  he  knows  himself. 

In  this  condition,  the  young  mechanic  applies  to  books 
to  assist  him  to  conquer  the  mysteries  of  his  Art;  but  he  has 
not  been  able  hitherto  to  find  a work  adapted  to  his  wants. 
He  anxiously  turns  the  pages  of  ponderous  quarto  and  folio 
volumes;  he  is  convinced  of  the  prodigious  learning  of  the 


10 


CARPENTRY  MADE  EASY. 


authors,  but  he  is  not  instructed  by  them.  On  the  one 
Hand,  their  practical  directions  and  rules  are  too  meagre ; 
and,  on  the  other  hand,  their  mathematical  reasoning  is  too 
technical  to  yield  our  young  working-man  any  real  benefit 
or  satisfaction.  May  not  these  faults  be  remedied?  Is  it 
not  possible  for  instruction  to  be  given,  which  shall  be  at 
once  simple  and  practical  in  detail,  and  comprehensible  and 
demonstrative  in  mathematical  reasoning? 

Design  of  this  Work. 

An  attempt  has  been  made,  in  this  little  work,  to  answer 
these  questions  affirmatively;  and  thus  to  supply  a positive 
want,  and  to  occupy  a new  field  in  the  literature  of  Archi- 
tecture. Its  design  is  to  give  plain  and  practical  rules  for 
attaining  a rapid  proficiency  in  the  Art  of  Carpentry ; and 
also  to  prove  the  correctness  of  these  rules  by  mathematical 
science. 

Importance  of  Geometry  to  Carpenters. 

No  certain  and  satisfactory  knowledge  of  framing  can  be 
gained  without  a previous  acquaintance  with  the  primary 
elements  of  Arithmetic  and  Geometry.  It  is  presumed  that 
a sufficient  knowledge  of  Arithmetic  is  possessed  by  most 
mechanics  in  this  country ; but  Geometry  is  not  so  commonly 
understood.  It  is  not  taught  in  our  District  Schools,  and  i? 
looked  upon  as  beyond  the  capacity  of  common  minds.  But 
this  is  a mistake.  To  mechanical  minds,  at  least,  the  ele- 
ments of  Plane  Geometry  are  so  easily  taught,  that  they 
seem  to  them  to  be  almost  self-evident  at  the  first  careful 
perusal;  and  mechanics  have  deprived  themselves  of  much 


INTRODUCTORY  CHAPTER. 


11 


pleasure,  as  well  as  profit,  in  not  having  made  themselves 
masters  of  this  science. 

Geometry  in  this  Work. 

Part  I.  is  therefore  devoted  to  so  much  of  the  Science  or 
Geometry  as  is  essential  to  the  complete  demonstration  and 
thorough  understanding  of  the  Science  and  Art  of  Carpentry; 
and  it  is  recommended  to  all  mechanics  into  whose  hands 
this  volume  may  fall,  to  give  their  days  and  nights  to  a 
careful  study  of  this  part  of  the  work.  It  is  true  that  our 
rules  and  instructions  in  Carpentry  are  so  plain  and  minute, 
that  they  are  available  to  those  who  do  not  care  to  study 
Geometry  at  all ; b.ut  the  principles  on  which  those  rules  are 
founded,  and  consequently  the  reasons  ivhy  the  rules  are  as  they 
are , cannot,  from  their  very  nature,  be  made  plain  and  in- 
telligible to  any  one  except  by  a course  of  geometrical  rea 
Boning. 

New  Rules  of  Carpentry. 

Part  II.  comprises  the  main  body  of  the  work,  and  is  de- 
voted particularly  to  the  framing  of  buildings.  The  rules 
for  obtaining  the  bevels  of  rafters,  joists,  braces,  &c.,  as  ex- 
plained in  this  part  of  the  work,  it  is  believed,  have  never 
been  published  before.  That  such  bevels  could  be  so  found 
has  been  known,  for  several  years  past,  among  master- 
builders  ; and,  to  a limited  extent,  has  by  that  means  been 
made  public;  but  this  feature  of  the  work  will,  no  doubt,  be 
new  and  useful  to  some  mechanics  who  have  followed  the 
business  for  years,  and  will  be  especially  useful  to  apprentices 
and  young  journeymen  who  have  not  yet  completed  their 
mechanical  education. 


12 


CARPENTRY  MADE  EASY. 


* 

They  are  Proved  and  Explained. 

These  rules  have  been  here  demonstrated  by  a new  and 
rigid  course  of  geometrical  reasoning;  so  that  their  correctness 
is  placed  beyond  doubt.  The  demonstrations  are  often  given 
in  foot-notes  and  in  smaller  print,  so  as  not  to  interrupt  the 
descriptive  portion  of  the  work,  nor  appall  those  who  are  not 
mechanically  learned,  by  an  imposing  display  of  scientific 
signs  and  technical  terms.  In  fact,  it  has  been  made  a lead- 
ing object,  in  the  preparation  of  this  work,  to  convey  correct 
mechanical  and  scientific  principles  in  simple  language, 
stripped  as  much  as  possible  of  all  technicalities,  and  adapted 
to  the  comprehension  of  plain  working-men. 

Bridge  Building. 

Part  III.  comprises  a brief  practical  treatise  on  the  framing 
and  construction  of  Bridges,  with  bills  of  timber  and  iron 
given  in  detail,  by  the  use  of  which  intelligent  carpenters 
can  construct  almost  any  kind  of  a bridge.  This  part  of  the 
wont  does  not,  however,  make  any  special  claims  to  new 
discoveries,  or  to  much  originality;  nor  is  it  intended  to 
supersede  the  use  of  those  works  specially  devoted  to  Bridge 
Building;  but  it  is  believed  it  will  be  found  more  practi- 
cally convenient  and  simple  than  some  others  of  more  im- 
posing bulk  and  of  higher  price. 

Valuable  Tables. 

Part  IV.  contains  a valuable  collection  of  Tables,  showing 
the  Lengths  of  Rafters,  Hip  Rafters,  Braces,  &c.,  and  also  the 
weights  of  iron,  the  strength  of  timber,  &c.,  &c.,  which  will 
be  found  of  the  greatest  convenience,  not  only  to  common 


INTRODUCTORY  CHAPTER. 


1 

mechanics  but  to  professional  designers,  architects,  and  bridge 
builders.  Some  of  these  tables  have  been  compiled  fron* 
♦reliable  sources;  but  the  most  important  of  them  have  beeu 
calculated  and  constructed,  at  a considerable  amount  of  ex 
pense  and  labor,  expressly  for  this  work. 

Plates  and  Illustrations. 

Nor  has  any  expense  been  spared  in  the  preparation  of  the 
plates  and  illustrations,  which  are  “got  up"  in  the  highest 
style  of  the  art ; and  it  is  hoped,  and  confidently  expected, 
that  the  work,  as  a whole,  will  prove  to  be  satisfactory  and 
remunerative  equally  to  the  Public  and  to  their 

Humble  and  obedient  servant, 

The  Author. 


The  Author  takes  great  pleasure  in  acknowledging 
the  eminent  services  rendered  him  in  the  literary  and 
scientific  portions  of  this  work,  by  E.  N.  Jencks,  A.  M., 
Professor  of  Mathematics  and  Natural  Sciences;  and 
the  Public  cannot  fail  to  appreciate  the  value  of  his 
labors  in  these  departments. 

The  inception  of  the  work,  its  original  designs,  and 
the  entire  system,  are  mine.  Whatever  is  found  in  it 
purely  literary  and  scientific,  I cheerfully  attribute  to 
his  assistance.  And  believing  that  the  work  will 
supply  a pressing  want,  and  will  be  useful  both  to 
those  who  are  devoted  to  the  Mechanic  Arts  and  to 
Amateurs  who  have  felt  the  necessity  of  a faithful 
guide  in  house-building  and  other  structures,  especially 
in  new  settlements,  I can  confidently  commend  it  to 
them  as  supplying  this  deficiency. 


PART  i. 


Plate  1. 


Plate  2 


GEOMETRY 


PLATES  I.  AND  II. 

Definitions. 

1.  Mathematics  is  the  science  of  quantity. 

2.  Quantity  is  any  thing  which  can  be  measured,  increased  or  diminished. 

3.  The  fundamental  Branches  of  Mathematics  are  Arithmetic  and 
Geometry.* 

4.  Arithmetic  is  the  science  of  numbers. 

5.  Geometry  is  the  science  of  magnitude. 

6.  Magnitude  has  three  dimensions : length,  breadth,  and  thickness. 

7.  A line  has  length  without  thickness.  The  extremities  of  a line  ar© 
called  points.  A point  has  no  magnitude,  but  position  only. 

8.  A straight  line  is  the  shortest  distance  between  two  points. 

9.  A curved  line  is  one  which  changes  its  direction  at  every  point.  It  is 
neither  straight  nor  composed  of  straight  lines. 

Thus  in  Eig.  1,  AB  is  a straight  line.  ACDB  is  a broken  line,  or  one 
composed  of  straight  lines  ; and  AFB  is  a curved  line. 

10.  The  single  term  line  is  often  used  in  the  sense  of  straight  line  ; and  the 
single  term  curve , of  curved  line. 

11.  Two  lines  are  parallel  when  they  are  everywhere  equally  distant. 

Fig.  2. 

12.  A surface  has  length  and  breadth  without  height  or  thickness. 

13.  A plane  is  a surface,  in  which,  if  any  two  of  its  points  be  joined  by  a 
straight  line,  that  line  will  lie  wholly  on  the  surface. 

14.  A solid , or  body , is  that  which  combines  the  three  dimensions  of  mag* 
nltudc,  having  length,  breadth,  and  thickness. 

15.  When  two  straight  lines  meet  each  other,  the  inclination  or  opening 

* Algebra  is  a branch  of  Mathematics,  but  can  scarcely  be  regarded  as  equally  fundament&J 
with  Arithmetic  and  Geometry. 

2 


(17) 


18 


CARPENTRY  MADE  EASY. 


between  them  is  called  an  angle  ; and  this  angle  is  said  to  be  greater  or  less 
as  the  lines  are  more  or  less  opened  or  inclined. 

The  vertex  of  an  angle  is  the  point  where  its  sides  meet.  Thus,  in  Fig. 
3,  A is  the  vertex,  and  AB  and  AC  are  the  sides. 

A ngles  occupy  surfaces  ; they  are  therefore  quantities  ; and  like  all  other 
quantities  are  susceptible  of  addition,  subtraction,  multiplication,  and  division. 
Thus,  in  Fig.  4,  the  angle  DCE  is  the  sum  of  the  two  angles  DCB  and  BCE 
A nd  the  angle  DCB  is  the  difference  of  the  two  angles  BCE  and  DCE. 

An  angle  is  designated  by  the  letter  at  the  vertex,  when  there  is  but  one 
angle  there,  as  the  angle  A in  Fig.  3 ; or  otherwise  by  the  three  letters  BAC 
or  CAB,  the  letter  at  the  vertex  being  always  placed  in  the  middle. 

16.  When  a line,  AB,  stands  on  another  line,  CD,  Fig.  5,  so  as  not  to  in- 
cline either  way,  AB  is  said  to  be  perpendicular  to  CD,  and  the  angle  on 
each  side  of  the  perpendicular  is  called  a right  angle. 

IT.  Every  angle  less  than  a right  angle  is  called  an  acute  angle,  as  DCB, 
in  Fig.  4 ; and  every  angle  greater  than  a right  angle,  as  ACD,  is  called  an 
obtuse  angle. 

18.  A polygon  is  a portion  of  a plane  terminated  on  all  sides  by  straight 
lines. 

19  An  equilateral  polygon  has  all  its  sides  equal,  and  an  equiangular 
polygon  has  all  its  angles  equal. 

20.  A regular  polygon  is  one  which  is  both  equilateral  and  equiangular. 

21.  The  polygon  of  three  sides  is  called  a triangle ; that  of  four  sides,  a 
quadrilateral ; one  of  five  sides,  a pentagon  ; one  of  six,  a hexagon;  one  of 
seven,  a heptagon;  one  of  eight,  an  octagon;  one  of  nine,  a nonagon;  one  of 
ten,  a decagon;  one  of  twelve,  a dodecagon;  one  of  fifteen,  a pent  e decagon ; 
and  so  on,  according  to  the  numerals  of  the  Greek  language. 

22.  An  equilateral  triangle  has  its  three  sides  equal : Fig.  6.  An  isosceles 
triangle  has  two  of  its  sides  equal.  A scalene  triangle  has  all  its  sides  un- 
equal. Figs.  T and  8. 

23.  A right-angled  triangle  contains  one  right  angle.  The  side  opposite 
the  right  angle  is  called  the  hypotenuse . Fig.  9 : AC,  opposite  the  right 
angle  B,  is  the  hypotenuse. 

24.  Quadrilaterals  a~e  designated  according  to  their  figures,  as  follows : 

The  square  has  its  sides  all  equal,  and  its  angles  all  right  angles.  Fig.  10 

The  rectangle,  or  oblong  square,  Fig.  11,  has  all  its  angles  right  angles 

and  its  opposite  sides  equal  and  parallel. 

The  parallelogram,  Fig.  12,  has  its  opposite  sides  equal  and  parallel. 
Every  rectangle  is  a parallelogram,  but  every  parallelogram  is  not  a rectangle. 

The  rhombus,  or  lozenge,  has  its  sides  all  equal  without  having  its  angles 
right  angles.  Fig.  13. 

The  trapezium  has  none  of  its  sides  parallel.  Fig.  14. 

The  trapezoid  has  two  of  its  sides  parallel.  Fig.  15. 


GEOMETRY. 


19 


25.  The  base  of  any  polygon  is  the  side  on  which  it  is  supposed  to  stand. 

26.  The  altitude  of  a triangle  is  the  perpendicular  let  fall  upon  the  base 
from  the  vertex  of  the  angle  opposite  the  base.  Thus,  in  Fig.  6,  AB  is  tho 
altitude  of  the  triangle  ACD. 

27.  The  altitude  of  a parallelogram , or  of  a trapezoid,  is  the  perpendicu 
lar  which  measures  the  distance  between  two  parallel  sides  taken  as  basest 
Thus,  in  Fig.  12,  AB  is  the  altitude  of  the  parallelogram  CD. 

28.  A diagonal  is  a line  within  a polygon,  which  joins  the  vertices  of  two 
angles  not  adjacent  to  each  other.  Thus,  in  Fig.  16,  AC,  AD,  and  AE  are 
diagonals. 

29.  The  area  of  a polygon  is  the  measure  of  its  surface. 

30  Equivalent  polygons  are  those  which  contain  equal  areas. 

31.  Equal  polygons  are  those  which  coincide  with  each  other  in  all  their 
parts.  (Ax.  13.) 

32.  Similar  polygons  have  the  angles  of  the  one  equal  to  the  angles  of  the 
other,  each  to  each,  and  the  sides  about  the  equal  angles  proportional. 

33.  Homologous  sides  and  homologous  angles  are  those  which  have  like 
positions  in  similar  polygons. 

34.  The  circumference  of  a circle  is  a curved  line , every  point  of  which 
is  equally  distant  from  a point  within,  called  the  centre. 

35.  The  circle  is  the  surface  bounded  by  the  circumference. 

36.  A radius  of  a circle  is  a straight  line  drawn  from  the  centre  to  the 
circumference.  (The  term  radius  is  a Latin  word,  the  plural  of  which  is 
radii.  Thus,  we  say  one  radius  and  two  radii.)  In  the  same  circle  all  radii 
are  equal ; all  diameters  are  also  equal,  and  each  diameter  is  double  the 
radius. 

37.  A diameter  of  a circle  is  a straight  line  drawn  through  the  centre,  and 
terminated  on  both  sides  by  the  circumference.  In  Fig.  17,  CD,  CG,  and 
CF  are  radii,  and  DE  is  a diameter. 

38.  An  arc  is  a portion  of  the  circumference ; as  AHB  in  Fig.  17. 

39.  A chord  is  the  straight  line  which  connects  the  two  extremities  of  an 
arc.  AB,  Fig.  17. 

40.  A segment  is  a portion  of  a circle  included  between  an  arc  and  its 
chor^  ; as  segment  AHB,  Fig.  17. 

41.  A sector  is  a portion  of  the  circle  included  between  two  radii;  ag 
sector  CGF,  Fig.  17. 

42.  An  inscribed  angle  is  one  formed  by  the  intersection  of  two  chords 
upon  the  circumference.  ABD  and  BDE  are  inscribed  angles. 

43.  An  inscribed  polygon  is  one  which,  like  EFG  in  Fig.  18,  has  all  its 
angles  in  the  circumference.  The  circle  is  then  said  to  circumscribe  such  a 
figure.  In  Fig.  17,  the  triangle  CGF  is  not  an  inscribed  triangle,  since  all 
the  angles  do  not  lie  in  the  circumference ; but,  in  the  same  Fig.,  DBE  is  an 
inscribed  triangle. 


20 


CARPENTRY  MADE  EASY. 


44  A secant  is  a line  which  intersects  the  circumference  in  two  points, 
and  lies  partly  within  and  partly  without  the  circle.  AB  is  a secant  in 
Fig.  It. 

45.  A tangent  is  a straight  line  touching  the  circumference  in  one  point  • 
only.  CD  is  a tangent — O is  the  point  of  contact. 

46.  The  circumference  of  every  circle  is  measured  by  being  supposed  to  be 
divided  into  360  equal  parts,  called  degrees ; each  degree  contains  60  win* 
ides,  and  each  minute  60  seconds.  Degrees,  minutes,  and  seconds,  are  desig- 
nated respectively  by  these  characters,  °,  ',  " ; thus,  45°  15'  30"  is  read  45 
degrees,  15  minutes,  and  30  seconds. 

4t.  Arcs  are  measured  by  the  number  of  degrees  which  they  contain. 
Thus,  in  Fig.  19,  the  arc  AE,  which  contains  90  degrees,  is  called  a quad- 
rant, or  the  quarter  of  a circumference,  because  90°  is  one  quarter  of  360°; 
and  the  arc  ACB,  which  contains  180°,  is  a semicircumference. 

48.  Every  angle  is  also  measured  by  degrees ; these  degrees  being  reck* 
oned  on  an  arc  included  between  its  sides,  described  from  the  vertex  of  the 
angle  as  a centre.  Thus,  in  Fig.  19,  the  right  angle  AOE  contains  90°; 
and  the  angle  BOD,  which  is  one  half  a right  angle,  is  called  an  angle  of  45 
degrees,  which  is  the  number  it  contains. 

49.  An  ellipse  is  a curved  line  drawn  around  two  points  within,  called 
foci*  (A,  B,  Fig.  20),  in  such  a manner  that  if,  from  any  point,  0,  of  the 
curve,  two  lines  be  drawn,  one  to  each  focus,  the  sum  of  these  two  lines,  AC 
and  BC,  shall  be  equal  to  the  sum  of  two  other  lines  drawn  to  the  foci  from 
any  other  point  of  the  curve ; as  DA  and  DB,  or  EA  and  EB. 

50.  The  centre  of  an  ellipse  is  the  middle  point  of  the  line  joining  the  two 
foci.  O in  Fig.  20. 

51.  The  diameter  of  an  ellipse  is  any  straight  line  passing  through  the 
centre,  and  terminated  on  both  sides  by  the  curve. 

52.  The  transverse  axis  of  an  ellipse  is  its  longest  diameter,  or  that  one 
which  passes  through  the  two  foci ; as  G I. 

53.  The  conjugate  axis  of  an  ellipse  is  its  shortest  diameter,  or  that  one 
which  is  perpendicular  to  the  transverse  axis;  as  DF. 

Note.  There  are  several  methods  employed  to  describe  an  ellipse,  but  the  one  which  is  at  once 
the  most  correct  and  practicable  is  by  means  of  the  instrument  called  the  Trammel,  represented 
in  Fig.  21.  It  consists  of  two  grooved  rules,  mitered  together  in  the  middle,  so  that  each  arm  id 
perfectly  perpendicular  to  the  two  adjacent  arms,  and  a rod  with  a movable  pencil  at  I*,  and  two 
Eaovable  pins,  one  at  A and  the  other  at  B. 

The  distance  from  B to  P equals  half  the  transverse  axis.  # 

The  distance  from  A to  P equals  half  the  conjugate  axis. 

The  distance  from  A to  B equals  half  the  distance  of  either  focus  from  the  center. 


• The  term  focus  is  a Latin  word,  of  which  the  plural  is  foci;  thus  we  say  one  focus  and  tw* 
foe « 


GEOMETRY. 


21 


Explanation  of  Mathematical  Symbols. 

In  order  to  facilitate  mathematical  calculations,  it  has  long  been  custom- 
ary among  civilized  nations  not  only  to  employ  figures  to  represent  numbers, 
but  also  to  employ  certain  other  signs  or  symbols  to  represent  such  opera- 
tions, in  the  combinations  of  numbers  and  quantities,  as  are  of  most  frequent 
occurrence ; and,  by  mutual  consent,  these  symbols  have  come  to  be  gener 
ally  known  and  employed  for  this  purpose. 

1.  The  sign  of  addition  is  written  thus  -j-,  and  is  read  plus;  for  example 
2 -j-3  is  read  two  plus  three,  and  signifies  two  added  to  three. 

2.  The  sign  of  subtraction  is  written  thus  — , and  is  read  minus  ;*  for  ex- 
ample, 3 — 2 is  read  three  minus  two,  and  signifies  three  less  two. 

3.  The  sign  of  multiplication  is  written  thus  X,  and  is  read  multiplied 
by ; thus,  3X2  is  read  three  multiplied  by.  two. 

4.  The  sign  of  division  is  written  thus  -i-,  and  is  read  divided  by , thus, 
12-7-3  signifies  12  divided  by  3.  Division  is  more  commonly  indicated,  how- 
ever, by  writing  the  divisor  under  the  dividend,  with  a line  between  them, 
in  the  form  of  a fraction ; thus,  Js2  signifies,  as  before,  12  divided  by  three. 

5.  The  sign  of  equality  is  written  thus  — , and  is  read  equals , or  is  equal 
to  ; for  example,  2+3=5  is  read  thus,  two  plus  three  equals  five. 

6.  The  letters  of  the  alphabet,  A,  B,  C,  &c.,  are  used  as  representatives 
of  quantities,  the  exact  dimensions  of  which  may  either  be  known  or  unknown. 
We  can  let  A,  for  example,  stand  for  a given  line,  a given  angle,  a given 
square,  or  a given  solid.  Lines  are  most  commonly  represented  by  the  two 
letters  placed  at  their  extremities ; and  angles  by  their  three  letters,  the  letter 
at  the  vertex  being  always  placed  in  the  middle. 

?.  A number  placed  before  a quantity  is  called  a co-efficient ; thus,  5AB 
is  read  five  AB,  or  five  times  AB,  or  AB  multiplied  by  five : the  sign  of  mul- 
tiplication being  understood  but  not  written. 

8.  A number  placed  at  the  right,  and  a little  above  a quantity,  is  called 
an  exponent , and  indicates  how  many  times  a quantity  is  taken  as  a factor; 
thus,  52  is  read  five  square;  it  is  equal  to  5X5,  and  signifies  that  five  is  to 
be  multiplied  by  five,  which  equals  25  ; also,  5*  is  read  five  cube  ; it  is  equal 
to  5X5X5,  and  signifies  that  five  is  to  be  multiplied  by  five,  and  that 
product  by  five,  which  equals  125. 

9.  This  sign  y/  is  used  to  show  that  a root  is  to  be  extracted.  A small 
figure  is  placed  in  the  bosom  of  the  sign,  called  the  index  of  the  root,  thus, 
y/  is  the  sign  of  the  square  root,  and  ■$/  is  the  sign  of  the  cube  root,  &c. 
When  no  index  is  written,  that  of  the  square  root  is  understood ; thus,  y/i 
represents  the  square  root  of  4. 


* Plus  and  minus  are  Latin  words,  the  former  meaning  more,  and  the  latter  I'ss;  these  words 
like  the  signs,  a^e  in  common  use  in  all  civilized  countries. 


22 


CARPENTRY  MADE  EASY. 


Definitions  of  Mathematical  Terms. 

1.  An  axiom  is  a self-evident  truth. 

2.  A theorem  is  a statement  which  requires  a demonstration,  by  reasoning 
from  such  truths  as  are  either  self-evident  or  previously  demonstrated. 

3.  A problem  is  a query  to  be  answered,  or  an  operation  to  be  performed. 

4.  The  term  proposition  may  be  applied  either  to  axioms,  theorems,  or 
problems. 

5.  A corollary  is  a necessary  inference  drawn  from  one  or  more  preceding 
propositions. 

6.  A scholium  is  an  explanatory  remark  on  one  or  more  preceding  propo- 
sitions. 

7.  An  hypothesis  is  a supposition  employed  either  in  the  statement  or  the 
demonstration  of  a proposition. 

8.  The  term  ratio  is  employed  to  denote  the  quotient  arising  from  dividing 
one  number  or  quantity  by  another:  for  example,  the  ratio  of  3 to  12  is  12  di- 
vided by  3 ; or ; ^ or  4.  The  ratio  can  always  be  expressed  in  the  form  of  a 
fraction,  whether  the  divisor  is  contained  in  the  dividend  an  exact  number  of 
times  or  not;  thus  the  ratio  of  2 to  1 is  J,  the  ratio  of  5 to  6 is  J ; and  so  also 

the  ratio  of  A to  B is  ?,  and  the  ratio  of  x to  y is  — . 

A x 

9.  Proportion  is  an  equality  of  ratios  or  an  equality  of  quotients  Thus 
when  the  quotient  arising  from  dividing  one  quantity  by  another  is  equal  to 
the  quotient  arising  from  dividing  a third  quantity  by  a fourth,  then  the 
four  quantities  are  said  to  be  in  proportion  to  each  other.  For  example,  the 
quotient  of  4 divided  by  2 equals  the  quotient  of  10  divided  by  5 ; or  | = l6°  ; 
then  these  four  numbers  2,  5,  4 and  10  are  in  proportion. 

Proportion  is  usually  indicated  by  writing  the  four  quantities  thus: 
2 : 4 : : 5 : 10,  and  is  read  2 is  to  4 as  fr  is  to  10  ; that  is,  2 is  just  such  a part 

of  4 as  5 is  of  10  ; for  2 is  half  of  4,  and  5 is  half  of  10.  So  also  if  ? = 

then  we  have  the  proportion  A : B : : C : D. 

10.  The  four  quantities  of  a proportion  are  called  its  terms.  The  first  and 
last  are  called  the  extremes , and  the  two  middle  ones  the  means  of  a propor- 
tion.. The  first  and  third  terms  are  called  the  antecedents,  and  the  second 
and  fourth  terms  are  called  the  consequents  of  a proportion. 

Axioms. 

1.  A whole  quantity  is  greater  than  any  of  its  parts. 

2.  A whole  quantity  is  equal  to  the  sum  of  all  its  parts. 

3 When  equals  are  added  to  equals,  their  sums  are  equal. 

4.  When  equals  are  added  to  unequals,  their  sums  are  unequal. 

6 WheL  equals  are  subtracted  from  equals,  their  remainders  are  equal 


GEOMETRY. 


23 


6.  When  equals  are  subtracted  from  unequals,  their  remainders  ajfc  unequal 

7.  When  equals  are  multiplied  by  equals,  their  products  are  equal. 

8.  When  equals  are  divided  by  equals,  their  quotients  are  equal. 

9 When  two  quantities  have,  each,  the  same  proportion  to  a third  quantity 
they  are  equal  to  each  other. 

10.  All  right  angles  are  equal. 

11.  When  a straight  line  is  perpendicular  to  one  of  two  parallels  it  is  per 
pendicular  to  the  other  also. 

12.  Only  one  straight  line  can  be  drawn  from  one  point  to  another. 

12  Two  magnitudes  are  equal,  when,  on  being  applied  to  each  other,  they 
coincide  throughout  their  whole  extent. 


Proposition  I.  Theorem. 


If  four  quantities  are  in  proportion , the  product  of  ike  two  means  will  equal 
the  product  of  the  two  extremes. 

Numerically.  Generally. 

Let  2 : 4 : : 5 : 10  ; A:B::C:D; 

then  will  4X5=2X10.  BXC=AXH 

For,  since  the  given  quantities  are  in  proportion,  their  ratios  are  equal 
(Def.  of  Terms,  9.) 

» i 4 10  B D 

And  we  have,  i=~. 


Multiply  both  quantities  by  the  divisor  of  the  first  ratio,  and  the  quantities 
will  still  be  equal  (Ax.  T)  ; we  shall  then  have, 


8Xj-8X$ 


Axa“Axc: 


or,  i=2X^.  B=AXg. 

Again,  multiply  both  quantities  by  the  divisor  of  the  second  ratio,  and  the  de 
sired  result  is  obtained  ; namely, 

4X5=2X10.  BXC=AXD. 


Proposition  II.  Theorem. 

When  the  product  of  two  quantities  equals  the  product  of  two  other  quanti- 
ties, then  two  of  them  are  the  means , and  the  other  two  the  extremes  of  a pro - 

portion . 

Numerically.  Generally. 

Let  4X5=2X10;  BXC=AXD; 

then  will  2 : 4 : : 5 : 10 ; A ; B : : C : D ; 

for,  divide  both  the  given  quantities  by  one  of  the  factors  of  the  first  quantity, 
which  will  not  alter  their  equality  (Ax.  8),  and  we  have, 

. 2X10  AXD 

4=“:  B=“c~i 


24 


CARPENTRY  MADE  EAS1. 


again,  divide  both  quantities  by  one  of  the  factors  of  the  second  quantity,  and 
we  have, 

4 _10  B_D 

2 5 ’ A“C* 

Here  we  have  an  equality  of  ratios,  and,  by  Def.  9,  the  four  quantities  are  in 
proportion ; hence, 

2:4  : :5  : 10.  A : B : : C : D. 

Scholium.  Quantities  are  said  to  be  in  proportion  by  inversion , when  the 
proportion  is  read  backward  ; 

thus,  4:2::  10:5;  B:A::D:C; 

or,  10  : 5 : : 4 : 2.  D : C : : B : A. 

Quantities  are  said  to  be  in  proportion  by  alternation,  when  they  are  read 
alternately ; 

thus,  2 : 5 : : 4 : 10  ; A : C : : B : D ; 

or,  4:  10::  2:5.  B, : D : : A : C. 

Quantities  are  said  to  be  in  proportion  by  composition,  when  the  sum  of  the 
antecedents  or  consequents  is  compared  with  either  antecedent  or  consequent 
thus,  2+5  : 5 : : 4+10  : 10  ; A+B  : B : : C+D  : D ; 

or,  2+5  : 2 : : 4+10  : 4.  A+B  : A : : C+D  : C. 


Proposition  III.  Theorem. 

When  four  quantities  are  in  proportion , they  will  also  he  in  proportion  by 


alternation. 

Let 

then  will 

for,  by  Prop.  I., 


Numerically. 

2 : 4 : : 5 : 10, 
2:5::  4: 10; 
2X10=5X4; 
and,  by  Prop.  II.,  2 : 5 : : 4 : 10. 


Generally. 

A : B : : C : D, 
A : C : : B : D ; 
AXD=CXB; 
A : C : : B : D. 


Proposition  IV.  Theorem. 

When  four  quantities  are  in  proportion , they  will  also  he  in  proportion 
inversion. 

Numerically.  Generally. 

Let  2 : 4 : : 5 : 10,  A : B : : C : D, 

then  will  10:5:  : 4 : 2 ; D:C::B:A; 

for,  by  Prop.  I.,  10X2=5X4;  DXA=CXB; 


and,  by  Prop.  II.,  10  : 5 : : 4 : 2. 


D : C : : B : A. 


GEOMETRY. 


25 


Proposition  V.  Theorem. 

When  there  are  four  proportional  quantities , and  four  other  proportional 
quantities,  having  the  antecedents  the  same  in  both,  the  consequents  will  be  pro 


portioned. 

Numerically. 

Generally. 

Let  2 : 4 : : 5 : 10,  and 

A :B::Cf:D, 

and  2 : 6 : : 5 : 15  ; 

A : X : : C : Y ; 

then  will  4 : 10  : : 6 : 15, 

B : D : : X : Y. 

Take  the  first  proportion  by  alternation  : 

2 : 5 : : 4 : 10  ; 

A : C : : B : D ; 

hence,  from  equality  of  ratios  (Def.), 

5 10 

C D 

2 4 * 

A B* 

Take  the  second  proportion  by  alternation 

: 

2 : 5 : : 6 : 15, 

A : C : : X : Y; 

and,  by  equality  of  ratios,  we  have, 

5_15. 

C Y 

2“"  6 ; 

A”X; 

, 10  15 

D Y 

hence,  1 

B“X; 

and  from  this  equality  of  ratios  there  results  (by  Def.), 

4 : 10  : : 6 : 15. 

B : D : : X : Y 

Corollary.  When  there  are  two  sets  of  proportional  quantities,  having  aa 
antecedent  and  a consequent  of  the  first  equal  to  an  antecedent  and  a conse^ 
quent  of  the  second,  the  remaining  quantities  are  proportional. 


Proposition  VI.  Theorem. 


When  four  quantities  are  in  proportion,  they  are  also  in  proportion  by  com- 
position. 


Numerically. 

Let  2 : 4 : : 5 : 1 0, 

then  will  2+4  : 2 : : 5+10  : 5. 

The  first  proportion  gives  (Prop.  I.), 


Generally. 

A : B : : C : D, 

A+B  : A : : C+D  : C. 


2X10=4X5.  AXD=BXC. 

Add  to  each  of  these  equal  quantities  the  product  of  the  two  antecedents, 
and  we  have, 

2X10  + 2X5=4X5+2X5;  AxD+AxC=BxC+A  XC ; 

or,  the  same  simplified, 

2X10+5=5X4+2;  AxD+C=CxB+A; 


hence,  by  Prop.  II., 

2+4  : 2 : : 5 + 10  • 5. 


A+B  : A : : C+D  : O. 


26 


CARPENTRY  MADE  EASY. 


Proposition  VII.  Theorem. 


If  any  two  quantities  be  each  multiplied  by  some  other  quantify,  t/teit 
products  will  have  the  same  ratio  as  the  quantities  themselves . 


Numerically. 

Let  2 and  4 

be  any  two  numbers ; 
multiply  each  by  5 ; 
then  2X5  : 4X5  : : 2 : 4 ; 

for,  (2X5)X4  = (4X5)X2, 

since  the  quantities  are  identical ; 
hence,  by  Prop.  II.,  2X5  : 4X5  : 


Generally. 

A and  B 

be  any  two  quantities ; 

S; 

AxS  : BxS  : : A : B ; 
(AXS)XB  = (BXS)XA, 

: 2 : 4.  AxS  : BXS  : : A : B. 


Proposition  VIII.  Theorem. 

When  two  triangles  have  two  sides  and  the  included  angle  of  the  one  equal  to 
two  sides  and  the  included  angle  of  the  other t each  to  each , the  two  triangles  art 
equal . 

In  the  triangles  ABC  and  DEF,  let  AB=DE,  AC=DF,  and  the  angle 
A = angle  D ; the  triangles  themselves  will  then  be  equal.  For,  apply  the  side 
AB  to  the  equal  side  DE,  so  that  the  point  A will  fall  upon  X),  and  the  point 
B upon  E ; then  since  angle  A=  angle  D,  the  side  AC  will  also  fall  upon  its 
equal  side  DF,  and  the  point  C upon  the  point  F; 
therefore  the  third  side,  CB,  will  fall  upon  the  third  side 
FE,  and  the  two  triangles  will  coincide  throughout 
their  whole  extent,  and  be  therefore  equal.  (Ax.  13.) 

Proposition  IX.  Theorem. 

When  two  triangles  have  two  angles  and  the  included 
side  of  the  one , equal  to  two  angles  and  the  included 
side  of  the  other , each  to  eacht  the  two  triangles  are 
equal . 

In  the  triangles  ABC  and  DEF,  let  the  angle 
A==  angle  D,  C = F,  and  the  included  side  AC  = DF  ; then  are  the  triangles 
also  equal. 

For,  apply  the  side  AC  to  its  equal  side  DF,  placing  the  point  A upon  the 
point  D,  and  the  point  C upon  F ; then,  since  the  angle  A=  angle  D,  the  side 
AB  will  take  the  direction  of  DE,  and  the  point  B will  fall  somewhere  upon 
the  line  DE  ; also,  since  the  angle  C=angle  F,  the  side  CB  will  take  the  di- 
rection of  FE,  and  the  point  B will  fall  somewhere  upon  the  line  FE  ; and 
since  the  point  B must  fall  upon  both  the  lines  DE  and  FE,  it  must  fall  upon 
E,  the  only  point  of  coincidence;  hence  the  two  triangles  coincide  through- 
out their  whole  extent,  and  are  therefore  equal.  (Ax.  13.) 

Cor  diary.  Every  triangle  has  six  parts,  namely  : three  sides  and  three 


GEOMETRY. 


27 


angles ; and  whenever  two  triangles  are  equal  to  each  other,  each  of  the  six 
parts  of  the  one  are  always  equal  to  the  corresponding  six  parts  of  the  other, 
side  to  side,  and  angle  to  angle.  It  is  to  be  observed,  also,  that  the  equal 
angles  are  always  opposite  to  the  equal  sides,  and  the  equal  sides  opposite  the 
equal  angles. 


../c 


A: 


Proposition  X.  Theorem. 

When  a straight  line  meets  another  straight  line,  the  sum  of  the  two  adjacent 
angles  are  equal  to  two  right  angles . 

Let  CD  meet  AB  at  D,  then  is  the  sum  of  the  two  angles  ADC  and  CDB 
equal  to  two  right  angles. 

Prom  the  point  D as  a centre,  describe  the  cir- 
cumference of  a circle,  then  will  the  line  AB  coin- 
cide with  its*  diameter,  since  it  passes  through  the 
centre.  (Def.)  Angles  are  measured  by  the  arcs 
intercepted  by  their  sides  (Def.)  ; and  since  the 
sides  of  the  angle  ADC  intercept  a portion  of  the 
semicircumference,  ABB,  and  the  sides  of  the  angle 
CDB  intercept  the  remaining  portion,  then,  both 
together  intercept  a semicircumference,  or  180  de- 
grees ; but  two  right  angles  intercept  180  degrees  (Def.)  ; therefore  the  sum  of 
the  two  angles  ADC  and  CDB=two  right  angles. 

Cor.  1.  When  one  of  the  given  angles  is  a 
right  angle,  the  other  is  a right  angle  also. 

Cor.  2.  When  one  line  is  perpendicular  to 
another,  then  is  the  second  line  also  perpendicular  to 
the  first. 

Let  CE  be  perpendicular  to  AB,  then  is  AB  per- 
pendicular to  CE. 

For,  since  CE  is  perpendicular  to  AB,  both  the 
angles  ADC  and  CDB  are  right  angles.  Again,  since  AD  is  a straight  line 
meeting  another  straight  line  CE  at  D,  then  the  sum  of  ADC-}-ADE=two 
right  angles  ; but  ADC  is  a 

angle  also.  Hence  AD,  or  AB,  is  perpendicular  to  CE. 

Cor.  3.  When  any  number  of  angles  have  their 
vertices  at  the  same  point,  and  lie  on  the  same  side 
of  a straight  line,  their  sum  is  equal  to  two  right  an 
gles,  for  they  all  together  intercept  an  arc  of  180G  aV 


right  angle  ; therefore  must  ADE  be  a right 


28 


CARPENTRY  MADE  EASY. 


Proposition  XI.  Theorem. 

The  opposite  or  vertical  angles,  formed  by  the  intersection  of  two  straight 
lines , are  equal. 

Let  AB  and  CD  be  two  straight  lines,  intersecting  c B 

each  other  at  E,  then  will  AEC=BED.  ^ 

For  the  6um  of  AEC+CEB  = two  right  angles 
(Prop.  X.)  ; and,  for  a similar  reason,  the  sum  of 
CEB+BED  = two  right  angles. 

Take  away  from  each  sum  the  common  angle  CEB,  and  there  remains 
AEC=BED. 

In  a similar  manner  it  may  be  proved  that  CEB=AED. 


Proposition  XII.  Theorem. 


If  two  parallel  straight  lines  meet  a third  line , the  sum  of  the  two  interior 
angles,  on  the  same  side  of  the  line  met , will  be  equal  to  two  right  angles. 

Let  the  two  parallel  lines  AB  and  CD  meet  the 
line  EF  ; then  will  BEF  + EFG=two  right  angles.  A v 
Through  E draw  EG,  perpendicular  to  CD,  and 
through  F draw  FH,  parallel  with  EG.  Then,  since 
parallels  are  everywhere  equally  distant,  (Def.  c , 

10),  we  have  EH=GF,  and  also  EG=HF ; 

and  since  AB  is  perpendicular  to  EG,  it  is  also  perpendicular  to  HP, 
(Ax.  11,)  and  the  angles  H and  G are  both  right  angles;  therefore,  the  two 
triangles,  EHF  andFGE,  are  equal,  (Prop.  VIII.)  And,  since  the  angles  op- 
posite the  equal  sides  are  equal,  (Prop.  IX.  Cor.),  angle  FEH=  angle 
EFG. 


But  the  sum  of  the  angles  BEF-f-FEH  is  equal  to  two  right  angles. 
(Prop.  X.)  Substitute  for  FEH  its  equal  EFG,  and  we  have  BEF-f  EFG= 
two  right  angles. 

Scholium.  Where  two  parallel  straight  lines  meet  a third  line,  the  angles 
thus  formed  take  particular  names,  as  follows  : 

Interior  angles  on  the  same  side  are  those  which 
lie  within  the  parallels,  and  on  the  same  side  of  the 
secant  line.  Thus  BEF  and  EFD  are  interior  an-  “ 
gles  on  the  same  side ; and  so  also  are  the  angles 
AEF  and  EFC.  a 

Alternate  angles  lie  within  the  parallels,  and  on  op- 
posite sides  of  the  secant  line,  but  not  adjacent  to 
each  other.  AEF  and  EFD  are  alternate  angles; 
also,  BEF  and  EFC. 

Alternate  exterior  angles  lie  without  the  parallels,  and  on  opposite  sides  of 
the  secant  line.  OEB  and  CFL  are  alternate  exterior  angles,  and  so  also 
are  AEO  and  LFD. 


GEOMETRY. 


29 


Opposite  exterior  and  interior  angles  lie  on  the  same  side  of  the  secant  line, 
the  one  without  and  the  other  within  the  parallels,  but  not  adjacent ; thus 
OEB  and  EFD  are  opposite  exterior  and  interior  angles ; so  also  are  BEF 
and  DFL. 

Cor.  1.  If  a straight  line  meet  two  parallel  lines,  the  alternate  angles  will 
be  equal.  For  the  sum  BEF -f EFD  = two  right  angles  ; also,  (by  Prop.  X), 
BEF+AEF=  two  right  angles  ; take  away  from  each  the  angle  BEF,  and 
there  remains  EFD=AEF. 

Cor.  2.  If  a straight  line  meet  two  parallel  lines,  the  opposite  exterioi 
and  interior  angles  will  be  equal.  For  the  sum  BEF-fEFD=  two  right  angles; 
also,  (by  Prop.  I.),  BEF+OEB=  two  right  angles ; taking  from  each  the 
angle  BEF,  and  there  remains  EFD  = OEB. 

Cor.  3.  Hence  of  the  eight  angles  formed  by  a line  cutting  two  parallel 
lines  obliquely,  the  four  acute  angles  are  equal  to  each  other,  and  so  also  are 
the  four  obtuse  angles. 

Proposition  XIII.  Theorem. 

If  two  straight  lines  meet  a third  line , making  the  sum  of  the  interior  angles 
on  the  same  side  equal  to  two  right  angles , the  two  lines  will  he  parallel . 

Let  the  two  lines  AB,  CD,  meet  the  third  line 
EF,  so  as  to  make  the  angles  BEF+EFG  equal  a h E,/j* 

to  two  right  angles;  then  will  AB  and  CD  be  pa-  j 

rallel,  or  everywhere  equally  distant.  j 

Through  E draw  EG  perpendicular  to  CD,  and  g — g i> 

through  F draw  FH  parallel  with  EG,  then  the  two 
angles  FEB-f-FEH  = two  right  angles,  (by  Prop. 

X.);  also,  the  angles  FEB-f  EFG==  two  right  angles,  by  hypothesis ; take  away 
from  each  the  angle  FEB,  and  there  remains  the  angle  FEH  — angle  EFG 
Again,  since  HF  and  EG  are  parallel  by  construction,  the  alternate  angles  EFH 
and  GEF  are  equal,  (by  last  Prop.,  Cor . 1)  ; hence,  the  two  triangles  EFH  and 
EFG  are  equal,  (Prop.  IX.),  having  two  angles  and  the  included  side  of  the  one 
equal  to  two  angles  and  the  included  side  of  the  other ; and  HF,  opposite  the 
angle  FEH,  is  equal  to  EG,  opposite  to  its  equal  angle  EFG.  (Prop.  IX., 
Cor.)  But  IIF  and  EG  measure  the  distance  of  the  line  CD  from  the  line 
AB,  at  the  points  II  and  E respectively.  The  same  demonstration  may  b< 
applied  to  any  other  two  points  of  the  line  AB ; hence  the  lines  AB  and 
CD  are  everywhere  equally  distant,  and  therefore  parallel. 

Cor.  1.  If  two  straight  lines  are  perpendicular  to  a third  line,  they  arc 
parallel  to  each  other;  for  the  two  interior  angles  on  the  same  side  are,  in 
that  case,  both  right  angles. 

Ccr.  2.  If  a straight  line  meet  two  other  straight  lines,  so  as  to  make 
the  alternate  angles  equal  to  each  other,  the  two  lines  will  be  parallel. 


80 


CARPENTRY  MADE  EASY. 


Let  OL  meet  AB  and  CD,  so  as  to  make  AEL= 

EFD  ; add  to  each  the  angle  BEF ; we  shall  then 
have  AEL  + BEF=EFD+BEF ; but  AEL-f  BEF=  i 
two  rightangles  (Prop.YIII.)  ; hence,  EFD-f-BEF= 
two  right  angles : therefore,  AB  and  CD  are  parallel. 

Cor.  3.  If  a straight  line,  OL,  meet  two  other  c ”75  p 

itraight  lines,  AB  and  CD,  so  as  to  make  the  ex-  / 
terior  angle,  OEB,  equal  to  the  interior  and  opposite 

angle,  EFD,  the  two  lines  will  be  parallel : for,  to  each  add  the  angle  BEF; 
we  shall  then  have  OEB-fBEF=EFD  + BEF  : but  OEB -f  BEF  are  equal 
to  two  right  angles:  therefore,  EFD  + BEF  is  equal  to  two  right  angles; 
and  AB  and  CD  are  parallel. 

Proposition  XIV,  Theorem. 

In  every  parallelogram,  the  opposite  angles  are  equal. 

Let  ABCD  be  a parallelogram  ; then  will  A=C, 
and  B=D. 

Draw  the  diagonal  BD ; then  will  the  triangle 
ADB=the  triangle  CBD  : for  the  angles  ABD  and 
BDC  are  alternate  angles  and  equal  (Prop.  XII., 

Cor.  1),  and  the  adjacent  sides,  AB=DC,  and  BD  is  common;  hence, 
the  triangles  are  equal  (Prop.  VIII.)  ; therefore  the  angles  A and  C,  oppo- 
site the  common  side  BD,  are  equal.  (Prop.  IX.,  Cor.)  In  a similar  manner 
it  may  be  proved  that  the  angles  B and  D are  equal. 

Cor.  1.  The  diagonal  of  a parallelogram  divides  it  into  two  equal  tri- 
angles. 

Cor.  2.  When  two  triangles  have  the  three  sides  of  the  one  equal  to  the 
three  sides  of  the  other,  the  angles  opposite  the  equal  sides  are  also  equal, 
and  the  triangles  themselves  are  equal. 

Cor.  3.  Two  parallels,  included  between  two  other  parallels,  are  equal. 

Cor.  4.  If  the  opposite  sides  of  a quadrilateral  are  equal,  each  to  each, 
the  equal  sides  will  also  be  parallel,  and  the  figure  will  be  a parallelogram ; 
for,  having  drawn  the  diagonal  BD,  the  triangles  ABD  and  BDC  are  equal; 
and  the  angle  ADB,  opposite  AB,  is  equal  to  the  angle  DBC,  opposite  DC. 
But  the  two  angles,  ADB  and  DBC  are  alternate  angles;  therefore,  AD  ia 
parallel  with  BC.  (Prop.  XIII.,  Cor.  2.)  ABD  and  BDC  are  also  equal  al- 
ternate angles;  therefore,  AB  is  parallel  with  DC,  and  the  figure  is  a paral- 
lelogram. 


Proposition  XV.  Theorem. 

When  two  angles  have  their  sides  parallel,  and  lying  in  the  same  direction, 
they  are  equal . 

Let  A BC  and  DEF  be  two  angles,  having  the  side  AB,  in  one,  parallel 


GEOMETRY. 


31 


to  DE,  in  the  other,  and  BC  parallel  to  EE,  and  lying  in  the  same  di- 
rection ; then  will  the  two  given  angles  be  equal. 

For,  produce  the  side  AB  till  it  intersects  EF  at 
G,  then  ABC=BGF,  for  they  are  opposite  in- 
terior and  exterior  angles  (Prop.,  XII.  Cor.  2)  ; 
also,  DEF  and  BGF  are  equal  for  a similar  rea- 
son : therefore,  ABC  and  DEF,  being  each  equal 
to  BGF,  are  equal  to  each  other.  (Ax.  9.) 


Proposition  XVI.  Theorem. 

The  sum  of  the  three  angles  of  any  triangle  is  equal  to  two  right  angles. 

Let  ABC  be  any  triangle.  Produce  the  base 
AB  to  any  convenient  distance,  as  D,  and  draw 
BE  parallel  with  AC  ; then  will  the  three 
angles  having  their  vertices  at  B be  equal  to  the  A 
three  angles  of  the  given  triangle,  for  the  angle  B,  or  ABC,  is  common  ; the  ai.gle 
c=C,  for  they  are  alternate  angles  ; and  the  angle  a= A,  for  they  are  opposite 
exterior  and  interior  angles.  But  the  sum  of  the  three  angles  at  B are  equal  to 
two  right  angles  (Prop.  X.,  Cor.  3)  ; hence  the  sum  of  the  three  angles  of 
the  given  triangle,  A-f-B  + C,  is  equal  to  two  right  angles. 

Cor.  1.  The  exterior  angle,  CBD,  of  any  triangle  formed  by  producing 
the  base,  is  equal  to  the  sum  of  the  two  opposite  interior  angles  of  the 
triangle. 

Cor.  2.  When  the  sum  of  two  angles  of  any  triangle  is  known,  the  third 
angle  is  found  by  subtracting  that  sum  from  two  right  angles  or  180°. 

Cor.  3.  When  two  angles  of  one  triangle  are  respectively  equal  to  two 
angles  of  another  triangle,  their  third  angles  are  also  equal,  and  the  triangles 
are  equiangular. 

Cor.  4.  It  is  impossible  3«jr  any  triangle  to  have  more  than  one  right 
angle,  for  if  it  could  have  two  right  angles,  the  third  angle  would  be  nothing. 
Still  less  can  any  triangle  have  more  than  one  obtuse  angle. 

Cor.  5.  In  every  right-angled  triangle,  the  sum  of  the  two  acute  angles  is 
equal  to  one  right  angle. 


Proposition  XVII.  Theorem. 

In  every  isosceles  triangley  the  angles  opposite  the  equal 
sides  are  equal. 

In  the  triangle  ABC  let  AC— BC  ; then  will  angle 
A=  angle  B.  Draw  the  line  CD  so  as  to  bisect  the  angle 
C,  that  is,  so  as  to  divide  it  into  two  equal  parts ; then 
are  the  two  triangles  ACD  and  BCD  equal  by  Prop. 
VIII.,  having  the  two  angles'  at  C and  the  two  adjacent 
sides  equal  Hence,  angle  A — B.  (Prop.  IX.,  Cor.) 


c 


32 


CARPENTRY  MADE  EASY. 


Cor.  1.  Every  equilateral  triangle  is  also  equiangular. 

Cor.  2.  The  equality  of  the  triangles  ADC,  and  BDC,  proves  that  the 
line  which  bisects  the  vertical  angle  of  an  isosceles  triangle  is  perpendicular  to 
the  base  at  its  middle  point,  for  the  two  angles  at  D are  each  right  angles. 
(Prop.  X.) 


Proposition  XVIII.  Theorem. 

When  (wo  angles  of  a triangle  are  equal , the  sides  op- 
posite them  are  also  equal , and  the  triangle  is  isosceles . 

Let  the  angle  A=B,  then  will  the  sides  AC  and  BC  be 
equal  also. 

Draw  CD  so  as  to  bisect  the  angle  C,  then  will  the  two 
triangles  be  equiangular  (Prop.  XVI.,  Cor.  3)  ; and  the 
side  CD  being  common,  the  two  triangles  are  equal  (Prop. 

IX.);  and  the  side  AC,  opposite  the  angle  B,  is  equal  to 
the  side  BC,  opposite  the  equal  angle  A. 

Proposition  XIX.  Theorem. 

Parallelograms  having  equal  bases  and  equal  altitudes , contain  equal  arcasi 
or  are  equivalent. 

Let  the  two  parallelograms,  ABCD 
and  ABEF,  have  the  same  base,  AB, 
and  the  same  altitude  PS  ; then  they  will 
be  equivalent. 

In  the  triangles  BCE  and  ADF,  the 
sides  BC  and  AD  are  equal,  being  opposite  sides  of  the  same  parallelogram; 
and  AF=BE  for  a similar  reason  ; the  included  angle  A is  equal  to  the  in- 
cluded angle  B,  since  their  sides  are  parallel  and  lie  in  the  same  direction 
(Prop.  XV.)  ; hence  the  two  triangles  are  equal.  (Prop.  VIII.) 

Now,  from  the  whole  quadrilateral  figure  ABCF,  take  away  the  triangle 
BCE,  and  there  remains  the  parallelogram  ABEF  ; from  the  same  quadri- 
lateral take  away  the  equal  triangle  ADF,  and  there  remains  the  parallelo- 
gram ABCD,  which  is  therefore  equivalent  to  ABEF. 


c 


Proposition  XX.  Theorem. 


Every  triangle  contains  half  the  area 
equal  altitude. 

Let  ABC  be  any  triangle,  and 
ADBE  be  a parallelogram  having 
the  same  base  and  altitude;  then  will 
the  triangle  contain  half  the  area  of 
the  parallelogram. 

Connect  C and  D,  and  complete  the  pa 


f a parallelogram  of  equal  base  ami 
cu  E 


p A E 


rallelogram  ADCF.  The  triangleBCF 


GEOMETRY. 


83 

is  half  the  parallelogram  FE,  (Prop.  XIY.,  Cor . 1)  ; and  the  triangle  ACF  is 
half  the  parallelogram  FD.  If  from  the  parallelogram  FE  we  take  the  paral- 
lelogram FD,  then  the  parallelogram  AE  will  remain  ; and  if  from  the  triangle 
BCF,  half  the  parallelogram  FE,  we  take  the  triangle  ACF,  half  the  parallel 
ogram  FD,  there  will  remain  the  triangle  ABC,  equal  to  one  half  the  paral- 
lelogram AE. 

Cor . 1.  The  demonstrations  in  this  and  the  preceding  propositions,  are 
equally  applicable  to  rectangles,  since  every  rectangle  is  also  a parallelogram  ; 
therefore,  every  rectangle  is  equivalent  to  a parallelogram  of  the  same  base 
and  altitude. 

Also,  every  triangle  is  equivalent  to  half  a rectangle  of  the  same  base 
and  altitude. 

Cor.  2.  Triangles  are  equivalent  to  each  other,  when  they  have  equal  bases 
and  equal  altitudes  ; each  being  half  an  equivalent  parallelogram. 

Proposition  XXL  Theorem. 

Two  rectangles  having  the  same  altitude  are  proportioned  to  each  other  as 
their  bases. 

Let  the  two  rectangles  AE  and  CF  have  equal 
altitudes,  then  will  their  surfaces  be  proportional  to 
the  length  of  their  bases.  a 

For,  since  their  altitudes  are  the  same,  and  their 
angles  are  all  right  angles,  they  may  be  so  applied 
to  each  other  that  the  whole  surface  of  the  shorter 
rectangle  shall  perfectly  coincide  with  an  equal  sur- 
face of  the  longer  one  ; and  this  coincidence  will  be  perfect  as  far  as  there  is 
a coincidence  of  their  bases,  and  no  further ; hence, 

AE  : CF  : : AB  : CD. 


Proposition  XXII.  Theorem. 

Rectangles  are  proportioned  to  each  other  as  the  products  of  their  bases  mul* 
tiplied,  by  their  altitudes. 

Let  P be  any  rectangle,  having  BC  for  its  base, 
and  BF  for  its  altitude ; and  let  N be  any  other 
rectangle,  having  AB  for  its  base,  and  BE  for  its 
altitude ; then, 

P : N : : BCxBF  : ABXBE. 

For,  place  the  two  rectangles  P and  N,  so  that 
the  base  AB  will  be  the  prolongation  of  the  base 
BC,  and  complete  the  rectangle  M;  then,  the  two  rectangles  P and  M,  hav- 
ing the  same  altitude  BF,  will  be  proportioned  to  each  other  as  their  bases, 
CB  and  AB  (Prop.  XXI.)  And,  for  the  same  reason,  the  two  rectangles  N 
3 


1 

! M 

P 

B C 

N 

E 

34 


CARPENTRY  MADE  EASY. 


and  M,  having  the  same  altitude  AB,  will  be  to  each  other  as  their  bases  BB 
and  BF  ; hence,  we  have  the  two  proportions  : 

P : M : : BC  : AB ; and 
M : N : : BF  : BE. 

Combining  these  two  proportions,  by  multiplying  the  corresponding  terms 
together,  w*t  have 

PXM  : NXM  : : BCxBF  : ABxBE 

But  the  quantity  M,  since  it  is  common  to  both  antecedent  and  consequent, 
can  be  omitted ; and  the  remaining  quantities  will  still  be  proportional. 
(Prop.  VII.)  Hence, 

P : N : : BCXBF  : ABXBE. 

Cor.  1.  Hence  the  area  or  surface  of  any  rectangle  is  measured  by  the 
product  of  its  base  multiplied  by  its  altitude  ; and  if  its  base  be  BC,  and  its 
altitude  BF,  its  area  or  measure  is  BCxBF. 

Cor.  2.  Since  the  sides  of  every  square  are  all  equal,  and  since  all  squares 
are  rectangles,  the  area  of  any  square  is  expressed  by  the  product  of  a side 
multiplied  by  itself : so  if  its  side  is  AB,  its  area  is  AB2. 

Cor.  3.  Since  every  rectangle  is  a parallelogram,  and  since  all  parallelo- 
grams of  the  same  base  and  altitude  are  equivalent,  (Prop.  XIX.),  there- 
fore the  area  of  any  parallelogram  is  the  product  of  its  base  by  its  altitude. 

Cor.  4.  Parallelograms  of  the  same  base  are  proportioned  to  each*  other 
as  their  altitudes,  and  those  of  the  same  altitude  as  their  bases ; and,  in  all 
cases,  they  are  proportioned  to  each  other,  as  the  products  of  their  bases  by 
their  altitudes. 

Proposition  XXIII.  Theorem. 

The  area  of  any  triangle  is  measured  by  the  product  of  its  base  multiplied 
by  half  its  altitude. 

Let  ABC  be  any  triangle,  of  which  AB  is  the 
base,  and  CD  the  altitude.  This  triangle  is  half  the 
parallelogram  AE,  (Prop.  XIV.,  Cor.)  ; but  the 
parallelogram  is  measured  by  its  base,  AB,  multi- 
plied by  its  altitude,  DC ; therefore  the  triangle  is  A 
measured  by  the  base  multiplied  by  half  the  altitude. 

Cor  Triangles  of  the  same  altitude  are  proportioned  to  each  other  as  theii 
bases,  and  those  of  the  same  bases  are  to  each  other  as  their  altitudes;  and, 
in  any  ^a«e,  they  are  proportioned  to  each  other  as  the  products  of  their 
bases  by  their  altitudes. 

Proposition  XXIV,,  Theorem. 

In  every  right-angled  triangle , the  square  of  the  hypotenuse  is  equal  to  iht 
rum  of  O'e  squares  of  the  other  two  sides. 

Let  ABC  be  a triangle,  having  the  angle  C a right  angle ; then  will 
AB*=AClJ-CB*. 


GEOMETRY. 


33 


Complete  the  squares  of  the  three  sides  of  the  given  triangle,  and  let  M 
represent  the  square  described  on  AB,  r c AB2;  let  N represent  the  square 
described  on  CB,  or  CB2;  and  let  P represent  the  square  described  on  AC, 
or  AC*.  Draw  the  diagonals  DB,  CE,  Cl,  and  AH,  and  from  C let  fall  CQ 
perpendicular  to  AB. 

In  the  two  triangles  DAB  and  /s 

CAE,  AC=AD,  each  being  a side  of  / \ 

the  square  P ; and  AB=AE,  each  be- 


ing a side  of  the  square  M ; the  in« 


/\ 
7 


0/ 


p 

/ 

fcs.  

!>r"  \ 
i \,  *"***V#‘ 

A!  / 

!F  \ !' 

! / 

\ | 

/ 

\ 

\ 

5 / 

\ ! 

1 

! M \ ! 

\ i 

IL.  , 

_ A 

/ 


also,  the  triangle  CAE  is  equivalent 


eluded  angle  DAB  is  made  up  of  tbe 
right  angle  DAC  and  the  angle  CAB  ; 
the  included  angle  CAE,  in  the  other 
triangle,  is  made  up  of  the  same  angl  > 

CAB,  and  the  right  angle  BAE ; 
hence,  the  angles  CAE  and  DAB  ar* 
equal,  and  the  triangles  themselves  are 
equal  fProp.  VIII.),  each  having  tw> 
sides  and  an  included  angle  equal. 

The  triangle  DAB  is  equivalent  to 
half  the  square  P,  for  it  has  the  same 
base  AD,  and  the  same  altitude  AC  ; 
to  half  the  rectangle  FAGE,  for  it  has  the  same  base  AE,  and  the  same 
altitude  AF ; hence,  the  rectangle  FAGE  is  equivalent  to  the  square  P„ 
Again,  the  two  triangles  ABH  rod  CBI  are  equal,  having  also  two  sides 
and  the  included  angle  of  the  one  equal  to  two  sides  and  the  included  angle 
of  the  other;  and  AHB  Is  half  of  ’he  square  N,  and  CBI  is  half  of  the  rect- 
angle FBIG  : therefore,  the  square  N is  equivalent  to  the  rectangle  FBIG. 

But  the  two  rectangles  FAGE  ar4  FBIG  make  up  the  square  M ; hence, 
M—P-j-N,  or  AB*=AC2-f  CB2. 

Cor.  1.  In  every  right-angled  t-iangle,  the  square  of  one  side  is  equiva- 
lent to  the  square  of  the  hypotenus  * less  the  square  of  the  other  side.  For 
example,  in  the  triangle  above,  AC*  =AB2— BC2;  also,  B C 2 — A Ba— A C*. 

Cor.  2.  Every  square  is  equal  to  half  the  square  of  its  own  diagonal. 

Let  AD  and  BC  be  the  diagonals  of  the  square 
ABCD  ; through  the  points  A and  D draw  straight 
lines  equal  and  parallel  with  BC ; and  through  the 
points  B and  C draw  lines  equal  and  parallel  with 
AI)  ; the  figure  thus  formed  will  be  ^he  square  of 
die  diagonal  CB,  or  of  its  equal  EF  : but  this  figure 
contains  eight  equal  triangles,  of  which  the  given 
Iquaie  contains  but  four;  hence, 

CB2 : AB2 : : 2 : 1 


v 


36 


CARPENTRY  MADE  EASY. 


and,  on  extracting  the  square  root  of  each  of  the  terms  of  this  proportion,  we 
have, 

CB  : AB  : : v/2  : 1 ; 

or,  the  diagonal  of  a square  is  proportioned  to  its  side,  as  the  square  roo\ 
of  two  is  to  one. 


Proposition  XXV.  Theorem. 

In  any  triangle , a line  drawn  parallel  to  the  base  divides  the  other  two  siilet 
proportionally. 

Let  ABC  be  any  triangle,  and  let  DE  be  parallel  with 
the  base  AB.  Draw  AE  and  BD.  The  two  triangles 
ADE  and  CDE,  having  the  same  altitude  DE,  are  in 
proportion  to  each  other  as  their  bases  AD  and  CD 
(Prop.  XXIII.,  Cor.)  ; also,  the  two  triangles  BED  and 
CED,  having  the  same  altitude  ED,  are  to  each  other  as 
their  bases  BE  and  CE  ; hence  the  two  proportions : a 

ADE : CDE : : AD : CD ; 

BED  : CDE  : : BE  : CE. 

The  two  triangles  ADE  and  BED  are  equivalent,  having  the  same  base, 
AB,  and  the  same  altitude,  since  the  line  DE  is  parallel  with  BC  ; hence,  the 
two  proportions  above  having  an  antecedent  and  a consequent  of  one  equiva- 
lent to  an  antecedent  and  a consequent  of  the  other,  the  remaining  terms  are 
proportional  (Prop.  V.,  Cor.)  ; hence, 

AD:  CD::  BE:  EC; 

and,  by  composition,  AD-fCD  : CD  : : BE-f  EC  : EC  ; 
or,  AC  : CD  : : BC  : CE  ; 

and,  by  alternation,  CD  : CE  : : AD  : BE. 

Proposition  XXVI.  Theorem. 

In  any  triangle , the  line  which  bisects  the  vertical  angle , when  produced  to 
the  base , divides  the  base  into  two  parts , which  are  proportional  to  the  adja • 
cent  sides. 

Let  ABC  be  any  triangle,  and  let  CE  bisect 
the  vertical  angle  C ; then  will 

BE  : BC  : : EA  : AC. 

The  angles  ACE  and  BCE  are  equal  by  hy- 
pothesis; draw  AD  parallel  with  CE,  and  pro- 
duce it  until  it  intersects  the  prolongation  of  BC 
at  D ; then  will  angle  D = angle  BCE  ; for  they 
are  opposite  exterior  and  interior  angles.  Also,  the  angle  DAC=ACI3, 
since  they  are  alternate  angles  ; hence,  those  two  angles  D and  A,  in  the  tri- 
ungle  C \ D,  equal  each  other,  and  the  triangle  is  isosceles.  (Prop.  XViJI  ) 


GEOMETRY. 


37 


In  the  triangle  BAD,  since  EC  is  parallel  with  the  base  AD,  it  divides  the 
other  two  sides  proportionally  (Prop.  XXV.),  and  we  have 

BE  : BC  : : EA  : CD  ; 

but  we  have  proved  the  triangle  CAD  to  be  isosceles;  hence,  AC=CD. 
Substitute,  therefore,  in  the  last  proportion,  AC  for  its  equal  CD,  and  we 
have,  BE  : BC  : : EA  : AC. 

Proposition  XXVII.  Theorem. 

All  equiangular  triangles  are  similar } and  have  their  homologous  sides  pro- 
portional. 

Let  ABC  and  DEA  be  two  triangles,  having  F 
the  angles,  C=E,  D=CAB,  and  B = DAE,  then  /^\NC 

will  their  homologous  sides  be  proportional,  and  e J 
we  shall  have  / N.  / 

BA  : AD  : : BC  : AE  ; and  n 

DA  : AB  ::  DE  : AC. 

Place  the  two  triangles  so  that  the  side  AD  shall  be  the  prolongation  of 
the  homologous  side  AB,  and  produce  DE  until  it  intersects  the  prolongation 
of  BC  at  F. 

Then  since  the  angles  EDA  and  CAB  are  equal,  the  lines  FD  and  CA  are 
parallel,  for  the  angles  are  opposite  exterior  and  interior  angles  (Prop.  XIII. 
Cor.) ; and  since  the  angles  DAE  and  ABC  are  equal,  the  lines  BF  and  AE 
are  parallel,  for  those  angles  are  opposite  exterior  and  interior  angles  also ; 
the  figure  ACEF  is  therefore  a parallelogram,  and  has  its  opposite  sides 
equal. 

In  the  triangle  BDF,  AC  being  parallel  with  the  base  DF,  the  other  two 
sides  are  divided  proportionally  (Prop.  XXV.)  ; 
and  we  have 

BA  : AD  : : BC  : CF.  But  AE=CF  ; hence, 

BA:  AD  ::  BC  : AE. 

In  the  same  manner  it  may  be  proved  that, 

DA  : AB  : : DE  : AC. 

Scholium.  It  is  to  be  observed,  that  the  homologous  or  proportional  sides 
are  opposite  to  the  equal  angles. 

Cor . Two  triangles  are  similar,  and  have  their  homologous  sides  propor- 
tional, when  two  angles  of  the  one  are  respectively  equal  to  two  angles  of  the 
other ; for  in  that  case  the  third  angles  must  also  be  equal  (Prop.  XVI. 
Cor,),  and  the  triangles  be  equiangular. 

Proposition  XXVIII.  Theorem, 

In  every  convex  polygon,  the  sum  of  the  interior  angles  is  equal  to  two  right 
angles , multiplied  by  the  number  of  sides  of  the  given  polygon,  less  two. 

Let  ABCDEF  be  any  convex  polygon,  and  let  diagonals  be  drawn  from 


38 


CARPENTRY  MADE  EASY. 


any  one  angle,  A,  to  each  of  the  other  angles  not  ad- 
jacent to  A ; these  diagonals  will  divide  the  polygon 
into  as  many  triangles,  less  two,  as  the  polygon  has 
sides,  whatever  the  number  of  the  sides  may  be. 

The  sum  of  the  angles  of  every  triangle  being  equal 
lo  two  right  angles  (Prop.  XVI.),  therefore  the  sum 
of  all  the  angles  of  the  given  polygon  will  be  equal  to 
twice  as  many  right  angles  as  there  are  triangles  thus  formed  within  it;  so 
that,  in  order  to  ascertain  the  entire  measure  of  the  angles  in  any  polygon,  we 
have  only  to  multiply  two  right  angles  by  the  number  of  its  sides  less  two. 

Cor . Since  2X2=4,  the  simplest  mode  of  estimating  the  measure  of  the 
angles  of  any  polygon,  is  to  multiply  the  entire  number  of  its  sides  by  two 
right  angles,  and  subtract  four  from  the  product. 

A quadrilateral  contains  four  right  angles,  since  4X2=8,  and  8 — 4=4. 

X pentagon  contains  six  right  angles,  since  5X2=10,  and  10—4=6. 

A hexagon  contains  eight  right  angles,  since  6X2=12,  and  12—4=8. 

A heptagon  contains  ten  right  angles,  since  1X2=14,  and  14 — 4=10. 

Proposition  XXIX.  Theorem. 

When  two  triangles  have  the  three  sides  of  the  onef  respectively  parallel  or  per- 
pendicular to  the  three  sides  of  the  other , the  two  triangles  are  similar. 

In  the  two  triangles  ABC  and  DEF,  c f 

First.  Let  the  sides  bo  respectively  parallel; 
namely,  let  AB  be  parallel  with  DE,  BC  with  EF, 
and  AC  with  DF ; then  the  angles  are  respectively 
equal;  namely,  A=D,  B=E,  and  C=F,  since  A 
their  sides  are  respectively  parallel  and  lying  in 
the  same  direction.  (Prop.  XV.)  Hence,  their 
homologous  sides  are  proportional,  and  they  are 
similar.  (Prop.  XXVII.) 

Secondly.  Let  the  sides  of  the  one  be  respectively  A 
perpendicular  tc  the  sides  of  the  other  ; namely, 

ED  perpendicular  to  AB,  FE  to  BC,  and  DF  to  AC ; then  they  will  still  be 
equiangular  and  similar. 

In  the  quadrilateral  LADI,  the  sum  of  the  four  interior  angles  is  equal  to 
four  right  angles  (Prop.  XXVIII.,  Cor.)  ; but  the  angles  L and  I are  each 
right  angles,  since  DL  is  given  perpendicular  to  AC,  and  ED  to  AB;  there- 
fore, the  sum  of  the  two  angles  A and  LDI  is  equal  to  two  right  angles; 
but  the  sum  of  the  angles  LDI  and  LDE  equals  two  right  angles  (Prop.  X.)  ; 
take  away  the  common  angle  LDI  from  each  sum,  and  there  remains,  A = 
LDE. 

For  similar  reasons,  B=DEF,  and  C = EFD;  hence,  the  two  triangles, 
being  equiangular,  have  their  homologous  sides  proportional,  and  are  similar 
(Prop.  XXVII.) 


c 


GEOMETRY, 


39 


Scholium.  The  homologous  sides  are  those  which  are  perj  endicular  or 
parallel  with  each  other,  since  they  are  also  those  which  lie  opposite  the  equa1 
angles. 

Proposition  XXX.  Problem. 

To  inscribe  a square  within  a given  circle. 

Let  ABCD  be  the  circumference  of  any  circle, 
and  let  two  diameters,  AC  and  BD,  be  drawn, 
intersecting  each  other  at  right  angles ; connect 
the  ends  of  these  diameters  by  the  chords  AB, 

BC,  CD,  and  DA,  then  will  these  chords  be 
equal  and  at  right  angles  with  each  other,  and 
thus  form  a perfect,  inscribed  square. 

For,  AO,  BO,  DO,  and  CO  are  all  radii  of 
the  same  circle,  and  therefore  equal  (Def.)  ; the 
four  angles  at  O are  right  angles  by  construc- 
tion ; hence,  the  four  triangles  AOB,  BOC,  COD,  and  DOA,  are  equal  (Prop. 
YIII.),  and  the  chords  opposite  the  equal  angles  at  0 are  also  equal.  (Prop. 
IX.,  Cor.) 

Again,  the  angles  BAD,  ADC,  DCB,  and  CBA  are  all  equal,  because  they 
are  each  composed  of  two  equal  angles  ; and,  since  their  sum  equals  four  right 
angles  (Prop.  XXYIII.,  Cor.),  each  one  is  a right  angle,  and  the  figure 
ABCD,  having  four  equal  sides  and  four  right  angles,  is  a square. 

Cor.  The  arcs  embraced  within  the  sides  of  the  equal  angles  at  0,  and 
intercepted  by  the  equal  chords,  are  all  equal,  since  each  one  is  the  fourth 
part  of  a circumference,  or  90°  ; hence,  in  the  same  circle,  or  in  equal  circles, 
equal  chords  intercept  equal  arcs,  and  equal  arcs  are  intercepted  by  equal 
chords. 


Proposition  XXXI.  Problem. 

To  inscribe  a regular  hexagon  and  an  equilateral  triangle  within  a given 
circle . 

Let  ABCDEF  be  the  circumference  of  any  a 

circle.  Draw  the  radii  AO  and  BO,  in  such 
a manner  that  the  chord  AB,  which  connects 
their  extremities,  shall  be  equal  to  the  radius 
itself.  This  chord  will  be  one  side  of  the 
regular,  inscribed  hexagon.  For,  the  triangle 
ABO,  being  equilateral,  is  also  equiangular 
(Prop.  XYII.,  Cor.))  and  the  sum  of  its 
three  angles,  being  equal  to  two  right  angles 
(Prop.  XYI.),  each  one  of  its  angles  is  equal 
to  tw?  thirds  of  a right  angle,  or  60°,  which 


40 


CARPENTRY  MADE  EASY. 


is  one  sixth  of  a circumference ; hence,  the  sides  of  the  angle  AOB  intercept 
one  sixth  of  the  circumference : therefore,  the  chord  AB,  applied  six  times  to 
the  circumference,  will  exactly  reach  around  it,  and  form  a regular  hexagon ; 
for  the  angles  of  this  hexagon  will  also  be  equal,  since  each  one  is  made  up 
of  two  equal  angles,  namely,  BAO-j-OAF  and  ABO-J-OBC,  &c. 

After  having  inscribed  the  regular  hexagon,  join  the  vertices  of  the  alter- 
nate angles  of  the  hexagon,  and  the  figure  thus  formed  will  be  an  equilateral 
triangle ; for  its  sides  are  chords  which  intercept  equal  arcs,  and  are  there- 
fore equal. 


PART  II. 


/■/"/ 


CARPENT  RY. 


PLATE  3. 

THE  USE  OE  THE  SQUARE  IN  OBTAINING  BEVELS. 

Although  the  square  is  one  of  the  first  instruments  placed  in  the 
hands  of  the  practical  carpenter,  yet  there  are  many  experienced  me- 
chanics who  have  never  learned  all  the  important  uses  to  which  it 
can  be  applied.  And  it  is  claimed  as  one  of  the  principal  merits  of 
this  work,  that  it  teaches  the  manner  of  obtaining  the  bevels  of  rafters, 
braces,  upper  joists,  gable-end  studding,  &c.,  in  the  most  simple  and 
most  accurate  manner  possible,  by  the  use  of  the  square  and  scratch 
awl  alone,  without  drafts  or  plans. 

The  Square  Described 

The  common  square  is  represented  in  Plate  3,  Fig.  1,  drawn  to  the 
scale  of  one  fourth  its  size.  The  point  0 is  called  the  cornel  oi  the 
heel  of  the  square,  the  part  OH  is  called  the  blade,  and  the  part  OP 
the  tongue.  The  blade  is  24  inches  long;  the  tongue  varies  in  length 
in  different  squares.  We  commence  at  the  heel  to  number  the 
inches  each  way. 


Pitch  of  the  Hoof. 

The  bevels  of  rafters,  joists,  &c.,  must,  of  course,  vary  with  the  pitch 
of  the  roof.  If  the  roof  is  designed  to  have  a quarter  pitch,  which  is 
the  most  common  inclination  for  a shingle  roof,  the  peak  of  the  roof 
will  be  a quarter  of  the  width  of  the  building  higher  than  the  top  of 
the  plates.  Although  this  is  called,  among  builders,  a quarter  pitch, 
yet  it  would  be  more  simple  to  call  it  a half  pitch  when  the  roof  has 
two  sides,  which  is  most  commonly  the  case,  for  the  true  inclination 

(43) 


44 


CARPENTRY  MADE  EASY. 


of  each  side  is  6 inches  rise  to  every  foot  in  width ; and  in  like  manner, 
a third  pitch  is,  in  reality,  a two-thirds  inclination  to  each  side  of  the 
roof,  for  it  has  8 inches  rise  to  every  foot  in  width. 

Bevels  of  Rafters. 

Let  AB;  in  Fig.  1,  represent  a rafter  which  is  required  to  be  beveled 
to  a quarter  pitch.  First  measure  the  exact  length  required,  upon  the 
edge  AB,  (which  will  be  the  upper  edge  of  the  rafter  when  it  assumes 
its  proper  place  in  the  frame,)  and  let  the  extreme  points  A and  B be 
marked.  Then  place  the  blade  of  the  square  upon  the  point  A at  the 
12  inch  mark,  and  let  the  tongue  rest  upon  the  edge  of  the  rafter  at 
the  6 inch  mark ; hold  the  square  firmly  in  this  position,  and  draw 
the  line  AC  along  the  blade:  this  line  will  be  the  lower  end  bevel, 
Take  the  square  to  the  other  end  of  the  rafter,  and  place  the  6 inch 
mark  on  the  tongue  upon  the  point  B,  still  having  the  blade  at  the 
12  inch  mark,  and  while  in  this  position  draw  the  line  DB  along  the 
tongue : this  line  will  be  the  upper  end  bevel  required. 

Proceed  in  a similar  manner  to  mark  the  bevels  for  any  other  pitch, 
placing  the  12  inch  mark  on  the  blade  upon  the  point  A,  and  that 
mark  on  the  tongue  which  corresponds  with  the  rise  of  the  roof  to 
the  foot,  on  the  point  B ; then  the  blade  of  the  square  will  show  the 
lower  end  bevel,  and  the  tongue  the  upper  end  bevel.  Thus,  if  the 
roof  has  a pitch  of  five  inches  to  the  foot,  let  the  square  be  placed  at  12 
and  5 ; if  the  roof  has  8 inches  rise  to  the  foot,  place  it  at  12  and  8,  &c. 

The  reason  of  this  rule  can  be  explained  in  few  words.  In  Fig.  2, 
let  C represent  the  middle  point  of  the  line  which  is  drawn  from  the 
top  of  one  plate  to  the  top  of  the  other;  let  AB  represent  a rafter;  and 
EC  the  longest  gable-end  stud,  having  its  longest  edge  EC  directly 
under  the  peak  of  the  roof  B.  The  lower  end  bevel  of  the  rafter  rests 
upon  the  upper  surface  of  the  plate,  which  is  horizontal  or  level,  while 
its  upper  end  bevel  is  perpendicular,  resting  against  the  upper  end  of 
the  opposite  rafter ; so  that  the  upper  and  lower  end  bevels  of  every 
rafter  are  always  at  right  angles  with  each  other,  whatever  the  pitch 
of  the  roof  may  be.  The  tongue  of  a square  is  also  always  at  right 
angles  with  the  blade  ; and  a square  can  be  conceived  as  having  its  heel 
at  the  point  C,  its  blade  resting  upon  the  line  AC,  and  its  tongue 
standing  perpendicularly  along  the  line  CB.  Now  let  the  distance 
from  A to  C be  supposed  to  be  1 foot,  or  12  inches ; then,  if  the  roof  is 
designed  to  rise  6 inches  to  the  foot,  the  point  B will  be  6 inches  from 
C ; ;f  it  rises  8 inches  to  the  foot,  the  point  B will  be  8 inches  from 


USE  OF  THE  SQUARE. 


45 


C,  &c. ; and  in  all  cases  the  line  AC  will  be  one  bevel,  and  the  hue  BC 
the  other. 

Bevels  of  Upper  Joists  and  Gable  End  Studding. 

The  bevel  of  the  upper  joists  is  always  the  same  as  the  lower  end  bevel 
of  the  rafters , and  the  bevel  of  the  gable  end  studding  is  the  same  as  the 
upper  end  bevel  of  the  rafters , whatever  the  pitch  of  the  roof  may  be.  For, 
in  respect  to  the  upper  joists,  it  is  to  be  observed  that  their  lower 
surfaces  rest  upon  the  plates,  and  their  ends  are  to  be  beveled  to  fit 
the  line  AB,  (Fig.  2),  hence  the  angle  BAD,  the  bevel  of  the  rafter, 
is  identical  with  that  of  the  end  of  the  joist.  The  bevel  of  the  end 
studding  is  designed  to  fit  it  to  the  lower  surface  of  the  rafter ; hence 
the  angle  DEC  is  the  proper  bevel.  But  DEC = ABC,  since  they  are 
opposite  exterior  and  interior  angles.  (Part  I.  Prop.  XII.,  Cor.) 

Bevels  of  Braces. 

Proceed  in  a similar  manner  to  obtain  the  bevels  of  braces.  When 
the  foot  and  the  head  of  the  brace  are  to  be  equally  distant  from  the 
intersection  of  the  two  timbers  required  to  be  braced,  and  when  the 
angle  of  their  intersection  is  a right  angle,  then  the  brace  is  said  to  be 
framed  on  a regular  run , and  the  bevels  will  be  the  same  at  both  ends 
of  it,  and  will  always  be  at  an  angle  of  45°,  which  is  half  a right 
angle,  or  the  eighth  part  of  a circle ; and  this  bevel  is  obtained  from 
the  square  by  taking  12  on  the  blade  and  12  on  the  tongue,  or  any 
other  identical  number,  the  rise  being  equal  to  the  run* 

But  when  the  foot  and  the  head  of  the  brace  are  to  be  at  unequal 
distances  from  the  intersection  of  the  timbers,  the  brace  is  said  to  be 
framed  on  an  irregular  run , and  the  bevel  at  one  end  will  be  different 
from  that  of  the  other.  One  rule,  however,  will  answer  in  all  cases. 
First  find  the  length  of  the  brace  from  the  extreme  point  of  one 
shoulder  to  the  extreme  point  of  the  other,  and  mark  those  points  as  A 
and  B.  Then  place  the  blade  of  the  square  upon  the  point  A at  such 
a distance  from  the  heel  as  corresponds  with  the  run  of  the  brace, 
while  the  tongue  crosses  the  edge  of  the  brace  at  that  distance  from  the 
heel  v\hicli  corresponds  with  the  rise  of  the  brace,  and  then  the  blade 
of  the  square  will  show  one  bevel,  and  the  tongue  the  other. 

For  example,  a brace  is  required  to  be  properly  beveled  for  an  ir- 


• For  the  explansition  of  those  terms,  rise,  run , &c.,  see  the  Introduction  to  the 
Tables.  Far  IV. 


46 


CARPENTRY  MADE  EASY. 


regular  run  of  4 by  5 feet.  Having  found  the  length  of  tbe  brace  (by 
Table  No.  4,  or  otherwise),  and  fixed  the  extreme  points  of  the 
shoulders,  then  lay  on  the  square  at  the  4 and  the  5 inch  marks,  and 
describe  the  bevels  along  the  blade  and  tongue  respectively,  as  in 
finding  the  bevels  of  rafters* 

Fig.  3 represents  a small  ivory  rule,  drawn  full  size.  It  is  intro- 
duced here  for  the  purpose  of  showing  the  manner  of  taking  the 
measure  of  hundredths  of  an  inch.  It  will  be  perceived  that  one  of  the 
inch  spaces  of  the  rule  is  divided  into  ten  parts,  by  lines  running 
down  diagonally  across  ten  other  horizontal  lines.  Each  of  the  in- 
tersections of  these  lines  measures  the  hundredth  part  of  an  inch ; the 
first  line  measuring  tenths,  the  second  twentieths,  ko. 


PLATE  4. 


Balloon  Frames. 


As  Balloon  Frames  are  the  simplest  of  all,  they  are  the  first  to  claim 
our  attention. 


The  Sills. 


Where  sqrare  timber  can  conveniently  be  furnished  for  sills,  it  is 
best  to  have  it ; but  small  buildings  can  be  very  well  constructed 
without  square  sills,  even  when  resting  upon  blocks  only,  by  using  a 
double  set  of  common  joists,  with  a 2 inch  space  between  them,  for 
the  tenons  of  the  studding. 

Such  a frame,  of  one  story  in  height,  16  feet  long,  and  12  feet  wide, 
is  represented  in  Plate  4.  For  this  building,  joists  which  are  2 
inches  thick  and  6 inches  wide,  will  answer.  First,  for  the  sills,  cut 
two  joists  16  feet  long,  and  two  others,  11  feet  8 inches  long.  Spike 
them  together  at  the  ends  in  the  form  of  an  oblong  square,  16  by  12 
feet,  making  the  outside  rim  of  the  sill. 


The  Studs. 

Next,  frame  18  studs  for  one  side  of  the  building;  the  two  comer 
studs  should  be  4 inches  square,  the  others  2 by  4.  Cut  out  a 2 inch 
relish,  six  inches  from  the  foot  of  each  stud,  on  the  face  side,  leaving 
a tenon  on  the  inside  of  6 inches  long  and  2 inches  square,  as  repre- 
sented in  Fig.  8.  Then  cut  off  the  other  end  of  the  studs  at  10  feet 
from  the  shoulder. 


The  Plates, 

A plate  of  2 by  4 stuff,  16  feet  long,  is  now  to  be  nailed  flat  upon  th© 
upper  ends  of  the  studs,  commencing  at  the  front  corner,  and  taking 
care  to  fix  them  14  inches  apart,  or  16  inches  from  centre  to  centre. 
The  last  space  will  often  be  more  or  less  than  14  inches ; but  it  is 
better  to  have  the  odd  space  all  at  one  end,  for  the  convenience  of 
the  plasterers  in  lathing. 

Raising  and  Plumbing  the  Frame. 

This  side  of  the  frame  is  now  ready  to  be  raised.  After  having 
prepared  the  other  side  in  the  same  manner,  that  can  also  be  raised 

(47) 


*8 


CARPENTRY  MADE  EASY. 


and  the  tenons  spiked  firmly  to  the  inside  of  the  sills.  The  corners 
should  then  be  plumbed  and  securely  braced. 

The  side  sills  should  now  be  completed  by  cutting  two*  joists,  one 
for  each  side,  each  15  feet  8 inches  long,  and  framing  them  for  the 
support  of  the  floor  joists  by  cutting  notches  into  their  upper  edges 
2 inches  wide  and  2 J inches  deep  ; cutting  the  first  notch  16  inches 
from  the  front  end,  and  the  next  one  just  14  inches  from  that,  and  so 
on  to  the  last.  After  these  inside  sill-pieces  are  thus  prepared,  they 
should  be  spiked  to  their  places  upon  the  inside  of  the  tenons  of  the 
studs. 


The  Floor  Joists. 

The  floor  joists  are  to  be  cut  off  11  feet  8 inches  long,  and  their  lower 
corners  notched  off  2 J inches  deep  ; then  they  should  be  fixed  in  their 
places  in  the  sills,  and  also  spiked  to  the  studs.  By  this  arrangement 
the  joists  are  left  one  inch  higher  than  the  sills  for  the  purpose  of 
having  the  door-sill  level  with  the  flooring;  the  door-sill  being  2 
inches,  and  the  flooring  1 inch  thick 

Upper  Joists. 

The  next  thing  is  to  frame  the  upper  joists,  the  rafters  and  the 
gable  end  studs ; beveling  the  ends  of  each,  so  as  to  correspond  with 
the  pitch  of  the  roof.  The  bevel  is  easily  found  by  the  use  of  the 
square,  as  is  explained  in  Plate  3.  The  upper  joists  are  equal  in 
length  to  the  width  of  the  building.  They  should  be  nailed  firmly 
upon  the  top  of  the  plates,  the  first  one  being  placed  4 inches  from 
the  end  of  the  plate,  to  leave  room  for  the  end  studding.  The  second 
one  should  be  14  inches  from  the  first,  and  the  others  at  the  same 
distance  from  each  other,  or  16  inches  from  centre  to  centre. 

The  Rafters. 

The  exact  length  of  the  rafters  is  found  by  the  use  of  Table  No.  1, 
Part  IY.  Look  at  the  left-hand  column  for  the  width  of  the  building, 
and  at  the  top  for  the  rise  of  the  rafter;  where  those  two  columns  meet 
in  the  table,  the  length  of  the  rafter  is  found  in  feet,  inches,  and  hun- 
dredths of  an  inch.  In  this  case,  the  width  of  the  building  is  12  feet, 
and  the  rise  of  the  rafter  is  6 inches  to  the  foot,  or  a quarter  pitch ; 
therefore  the  length  of  the  rafter,  as  given  in  the  table,  is  (6 : 8.49) 
6 feet  8 inches  and  ,Vo  of  an  inch.  The  rule  for  obtaining  these  lengths 
is  of  perfect  accuracy,  and  is  explained  in  the  introduction  to  the  Tables 


BALLOON  FRAMING. 


4U 


m such  a manner  that  every  carpenter  can  calculate  these  lengths  for 
himself,  from  the  primary  elements,  if  he  chooses.  The  size  of  these 
rafters  is  2 by  4. 

Gable-End  Studs. 

The  length  of  the  gable-end  studding  may  be  found  by  first  cal- 
culating the  length  of  the  longest  one,  which  stands  under  the  very 
peak,  and  then  obtaining  the  lengths  of  the  others  from  this ; or,  by 
first  calculating  the  length  of  the  shortest  one  next  to  the  corner  of 
the  building,  and  then  obtaining  the  lengths  of  the  others  from  this. 

The  length  of  the  middle  stud  is  found  by  adding  to  the  length  of 
the  side  studding,  the  rise  of  the  roof  and  the  thickness  of  the  plate  ; 
and  deducting  from  that  sum  the  thickness  of  the  rafter,  measured  on 
the  upper  end  bevel.  For  example,  in  this  building,  the  length  of  the 
side  studding  from  shoulder  to  shoulder  is  10  feet,  the  rise  of  the  roof  is 
3 feet,  and  the  thickness  of  the  plate  is  2 inches.  These  all  added  are 
13  feet  2 inches,  from  which  deduct  4.47  inches,  or  4 J inches,  the  thick 
ness  of  the  rafter  measured  on  the  upper  end  bevel,  and  the  result  is 
12  feet  10J  inches,  the  length  of  the  middle  stud.  The  next  stud,  if 
placed  16  inches  from  this  one,  from,  centre  to  centre,  is  8 inches 
shorter,  since  the  rise  of  the  roof  is  6 inches  to  12,  or  8 inches  to  16. 
The  next  one  is  8 inches  shorter  still,  and  the  others  in  proportion. 

If  it  should  be  thought  preferable  to  commence  by  first  calculating 
the  length  of  the  shortest  one,  it  can  be  done.  For  example,  in  this 
building  the  distance  of  the  inside  of  the  first  stud  from  the  outside  of 
the  building  is  20  inches,  the  rise  of  the  rafter  in  running  20  inches 
back  is  10  inches,  to  which  add  2 inches,  thickness  of  the  plate,  and 
10  feet  for  the  length  of  the  side  studding,  and  the  sum  is  11  feet; 
from  this  deduct  the  thickness  of  the  rafter,  at  the  upper  end  bevel, 
4J  inches,  and  the  result  is  10  feet  7|  inches,  the  length  of  the  shortest 
stud.  The  length  of  the  next  one  is  found  by  adding  the  rise  of  the 
roof  in  running  the  distance,  that  is,  if  they  are  16  inches  apart  from 
centre  to  centre,  the  difference  between  them  is  8 inches ; and  so  in 
any  other  pitch,  in  the  proportion  of  the  rise  to  the  run . 

The  end  studding  having  been  properly  beveled  and  cut  off  to  the 
exact  lengths  required,  they  can  be  raised  singly  and  spiked  to  the  sills 
at  the  bottom  and  nailed  at  the  top  to  the  end  rafters,  and  also  to  the 
upper  joists  where  they  intersect  them.  After  the  end  studs  are  all 
fixed  in  their  positions,  the  end  sills  can  finally  be  completed  by 
spiking  a joist  11  feet  in  length  to  the  inside  of  the  studding  at  each 
end  of  the  frame, 


PLATE  5. 


Plate  5 is  designed  to  represent  a balloon  frame  of  a building  a 
story  and  a half  high,  16  by  26  feet  on  the  ground,  with  12  feet  stud- 
ding. Two  end  elevations  are  given,  in  order  to  exhibit  different  styles 
of  roofs.  Fig.  2 being  a plain  roof,  of  a quarter  pitch ; and  Fig.  3 a 
Gothic  roof,  the  rafters  rising  14  inches  to  the  foot. 

Framing  the  Sills. 

Solid  timber,  8 inches  square,  being  furnished  for  the  sills  of  this 
building,  the  first  business  is  to  frame  these.  The  carpenter  will  seldom 
have  timber  furnished  to  his  hand  which  is  perfectly  square  through- 
out its  length;  by  carelessness  in  hewing,  or  by  the  process  of  season 
ing  after  being  hewed,  it  will  most  commonly  have  become  irregular 
and  winding. 

Work  Sides. 

Having  first  selected  the  two  best  adjoining  sides,  one  for  the  upper 
side  and  the  other  for  the  front,  called  work  sides , they  should  be  taken 
out  of  wind  in  the  following  manner. 

To  take  Timber  out  of  Wind. 

Plane  off  a spot  on  one  of  the  work  sides,  a few  inches  from  one  end* 
and  draw  a pencil  line  square  across  it ; then  place  the  blade  of  a 
square  upon  this  line,  allowing  the  tongue  to  hang  down  as  a plummet, 
to  keep  the  blade  on  its  edge.  Leave  the  square  in  this  position,  and 
go  to  the  other  end  of  the  sill,  and  place  another  square  upon  it,  in  the 
same  manner ; then  sight  across  the  two  squares,  and  see  if  they  are 
level  or  parallel  with  each  other.  If  not,  make  them  so,  by  cutting 
off  the  spots  under  the  squares  till  they  become  so ; then  make  the 
other  work  side  square  with  this  one,  at  these  two  spots,  and  draw  a 
pencil  mark  square  across  both  sides : these  marks  are  called 
plumb  spots. 

On  the  upper  side  of  the  timber,  strike  a chalk-line,  from  one  end 
to  the  other,  at  two  inches  from  the  front  edge;  this  will  be  the  front 
line  for  moitices  for  stuls.  On  this  line  measure  the  length  of  the 
(50) 


BALLOON  FRAMING. 


51 


sills,  and  square  the  ends  by  it.  If  the  stick  is  very  irregular,  it  should 
be  counter-hewed,  and  the  two  work  sides  made  square  and  straight. 

Spacing  for  Windows  and  Doors. 

Next,  lay  out  spaces  for  windows  and  doors,  leaving  a space  for  th@ 
doorway  2 J or  3 inches  more  than  the  width  of  the  door ; and  leave 
spaces,  7 inches  more  than  the  width  of  the  glass,  for  the  windows. 

Mortices  for  the  Studs. 

Then  lay  out  the  mortices  for  the  studding,  spacing  them  as  do* 
scribed  in  Plate  4.  The  studding  on  each  side  of  the  doors  and  win- 
dows should  be  4 inches  square,  as  well  as  those  at  the  corners  of  the 
building.  The  rest  of  the  studding  may  be  2X4.  The  mortices  need 
to  be  a little  more  than  2 inches  deep,  and  the  tenons  2 inches  long. 

The  lower  joists  for  this  building  should  be  2X8,  and  10  inches 
shorter  than  the  width  of  the  building.  They  should  be  placed  Id 
inches  apart,  from  centre  to  centre,  as  already  described. 

The  Gains, 

As  they  are  called,  for  receiving  the  ends  of  the  joists,  should  be 
cut  out  of  the  side  sills,  4 inches  deep  and  2 inches  square,  and  5 
inches  from  the  front  or  outside  of  the  sill.  Having  framed  the 
sills  for  the  studding  and  joists,  they  should  next  be  framed  for  each 
other.  Make  mortices  in  the  ends  of  the  side  sills,  2|  inches  from  the 
upper  surface  and  2 inches  from  the  end,  2 inches  wide  and  5J  inches 
long.  The  inside  of  the  sills  should  be  faced  off,  along  the  mortice^ 
to  within  7|  inches  of  the  work  side,  in  front.  The  length  of  the  side 
sills  should  be  the  same,  of  course,  as  that  of  the  building ; but  the 
end  sills  should  be  measured  from  shoulder  to  shoulder,  15  inches  less 
than  the  width  of  the  building.  Make  the  tenons  of  these  to  corres- 
pond with  the  mortices  of  those  which  have  just  been  described. 

The  Draw  Bores. 

The  draw  bores  should  be  1 inch  in  diameter,  and  1|  inches  from 
the  face  of  the  mortice.  The  draw  bore  through  the  tenon  should  b© 

of  an  inch  nearer  the  shoulder  than  that  through  the  mortice,  m 
order  to  draw  the  work  snugly  together. 

A Draw  Pin. 

The  proper  way  to  make  a draw  pin  for  an  inch  bore  is,  first,  to 


62 


CARPENTRY  MADE  EA3Y. 


make  it  an  inch  square ; then  cut  off  the  corners,  making  it  eight* 
square,  then  taper  it  to  a point,  the  taper  extending  one  third  the  length 
of  the  pin.  The  pin  should  be  about  2 inches  longer  than  the  thickness 
of  the  timber. 

The  sills  haying  thus  been  framed,  they  can  be  brought  to  their 
places  and  pinned  together,  and  then  the  lower  joists  laid  down. 

To  Support  the  Upper  Joists* 

This  building  being  a story  and  a half  high,  the  upper  joists  are 
laid  upon  a piece  of  inch  board,  from  4 to  6 inches  wide,  which  is  let 
into  the  studs,  as  seen  in  the  Plate.  The  bevels  and  lengths  of  the 
rafters  are  found  as  already  described. 

In  Fig.  3 the  rafters  are  represented  as  projecting  beyond  the 
plate ; this  projection  may  be  3 feet,  or  more,  according  to  each  one’s 
fancy ; but  whatever  it  may  be,  it  must  be  added  to  the  length  of  the 
rafter  as  given  in  the  table,  where  it  is  calculated  from  the  upper  and 
outer  corner  of  the  plate.  The  bevel  will  be  the  same,  whatever  the 
additional  length  may  be,  as  if  the  rafter  did  not  project  at  all.  In 
this  case,  the  rafter  should  be  cut  out  to  about  one  half  its  width, 
where  it  intersects  the  plate,  and  must  be  spiked  securely  to  the  plate. 
The  two  bevels,  at  the  intersection,  will  be  the  same  as  the  upper  and 
lower  end  bevels,  and  will  make  a right  angle  with  each  other  where 
they  meet  at  this  place. 

The  collar  beams  can  be  spiked  to  the  rafters,  or  they  can  be  dove- 
tailed into  them.  Both  methods  are  represented  in  the  plate. 


Plate  6 

FigJ. 


T 


PLATE  6. 


Plat©  No.  6 represents  a balloon  frame  of  a two-story  building 
18  by  80  feet,  with  18  feet  studding,  to  be  erected  upon  a good  stone 
or  brick  wall.  Heavy  joists,  8 by  10,  are  used  for  sills,  with  the  ends 
halved  together,  and  fastened  with  spikes,  as  represented  in  the  Plate. 
The  lower  joists  should  be  2 by  9 inches,  of  full  length,  equal  to  the 
width  of  the  building.  The  lower  corners  are  notched  off  8 inches, 
and  they  are  spiked  to  the  studding.  The  mortices  for  the  studs  should 
be  1J  inches  deep,  the  studding  being  2 by  4.  The  middle  joists  are 
2 by  9,  and  arranged  as  in  Plate  5 ; and  the  upper,  2 by  7,  and  ar- 
ranged as  in  Plate  4. 

Crowning  of  Joists. 

It  will  almost  always  happen  that  one  edge  of  a joist  will  have  be- 
come somewhat  rounded  out,  and  the  other  edge  rounded  in,  by  the 
process  of  seasoning ; and  it  is  of  much  importance,  especially  in  long 
joists  of  18  feet  or  more,  to  be  careful,  in  placing  the  joists  in  a 
building,  to  place  the  rounding  or  crowning  side  up. 

Bridging  of  Joists. 

Joists  12  feet  long,  or  over,  should  also  be  bridged  in  one  or  more 
places,  by  nailing  short  pieces  of  board,  2 or  3 inches  wide,  in  the  form 
of  a brace,  from  the  lower  edge  of  one  joist  to  the  upper  edge  of  the 
next  one ; and  then  another  piece,  from  the  lower  edge  of  this  one  to 
the  upper  edge  of  the  first  one;  and  so  on,  throughout  the  whole 
length  of  the  building : having  two  braces  crossing  each  other  between 
each  joist,  beveling  the  ends  so  as  exactly  to  fit,  which  would  add  very 
much  to  the  strength  of  the  floor. 

Lining  or  Sheeting  Balloon  Frames. 

After  an  experience  of  fifteen  years  in  constructing  and  repairing 
balloon- framed  buildings,  I have  found  it  best  to  line  the  frame  on 
the  inside  for  three  reasons : 

First — the  work  is  more  durable . For,  when  a frame  is  lined  on 
the  outside,  (the  common  way,)  it  is  very  difficult  to  weather-board 
it  sufficiently  tight,  to  prevent  the  rain  beating  in  between  the  siding 

(53) 


64 


CARPENTRY  MADE  EASY. 


and  the  lining,  and  thus  rotting  both,  since  there  is  so  little  opportu- 
nity there  for  the  moisture  to  dry  out. 

Second — the  lining  is  stiffer  and  warmer . For,  in  that  case,  the  lath 
being  but  half  an  inch  from  the  lining-boards,*  the  mortar  is  pressed 
in  between  every  board,  making  it  almost  air-tight. 

Third — the  wall  itself  is  made  more  solid.  For  the  mortar  being 
pressed  against  the  lining-boards,  is  forced  both  ways,  both  up  and 
down,  forming  more  perfect  clinchers. 


* When  a building  is  thus  lined  on  the  inside,  it  is  best  to  lath  it  in  the  following 
manner.  Single  strips  of  lath  are  first  nailed  perpendicularly,  sixteen  inches  apar^ 
upon  the  lining-boards,  and  to  these  the  laths  for  the  wall  are  nailed  as  usual 


PLATE  7. 


BARN  FRAMES. 

Plate  7 represents  the  frame  of  a barn  30  by  40  feet,  and  16  feet 
high  between  shoulders. 

The  sills  are  12  inches  square; 

Posts  and  large  girders,  10  inches  square ; 

Plates  and  girders  over  main  doors,  8 by  10 ; 

Purlin  plates,  6 by  6 ; 

Purlin  posts  and  small  girders,  6 by  8 ; 

Braces,  4 by  4 ; and  rafters,  2 by  6. 

First  proceed  to  take  the  timber  out  of  wind,  as  directed  under 
Plate  5.  Frame  the  sills  together  as  represented  in  the  Plate,  the 
four  short  sills  being  framed  into  the  two  long  ones,  having  taken 
care  to  select  the  best  of  the  short  sills  for  the  ends. 

Size  of  Mortices. 

The  mortices  for  the  end  sills  should  be  3 by  9 inches,  with  a relish 
of  2 1 or  3 inches  on  the  outside.  The  mortices  for  the  middle 
sills  may  be  3 by  11  inches.  The  mortices  for  the  corner  posts 
should  be  3 by  7 inches,  and  for  the  middle  posts,  3 by  9 inches;  all 
the  mortices  in  the  sills  being  3 inches  from  the  work  sides.  The 
general  rule  for  draw  bores  and  draw  pins  may  be  stated  as  follows : — 
The  size  of  the  draw  bore  should  be  equal  to  half  the  thickness  of  the 
tenon,  when  the  tenon  is  not  more  than  3 inches  thick ; but  it  never 
need  be  more  than  1J  inches  in  size,  even  though  the  tenon  may  be 
more  than  3 inches  thick.  In  wide  mortices,  it  is  customary  to  have 
the  tenons  secured  with  two,  and  sometimes  three  pins,  as  represented 
in  the  Plate.  Let  one  draw  bore  be  2 inches  from  one  side  of  the 
mortice,  and  the  other  2 inches  from  the  other  side,  and  each  one  2 
inches  from  the  face  of  the  mortice. 

In  the  tenons,  let  the  draw  bores  be  2 inches  from  each  side,  and 
about  one  fourth  of  an  inch,  in  large  tenons,  nearer  the  shoulder  than 
the  draw  bores  of  the  mortices.  Great  care  should  be  observed  to 
have  the  draw  bores  perfectly  plumb ; and  workmen  should  be  cau- 
tioned against  making  a push  horey  as  it  is  called,  when  not  plumb. 

(55) 


56 


CARPENTRY  MADE  EASY. 


The  posts  need  not  be  pinned  at  the  bottom,  and  the  manner  of  pin- 
ning the  other  tenons  is  represented  in  the  Plate. 

Braces. 

The  braces  are  framed  on  a regular  3 feet  run ; that  is,  the  brace 
mortice  in  the  girder  is  3 feet  from  the  shoulder  of  the  girder,  and  the 
brace  mortice  in  the  post  is  3 feet  below  the  girder  mortice.  AU 
ways  remember  that  the  measure  for  braces  and  brace  mortices  is  com- 
puted to  the  furthest  end,  or  toe  of  the  brace , and  the  furthest  end  of 
the  mortice.  The  mortices  for  4 inch  braces  need  to  be  5 J inches 
long,  so  that  the  end  of  the  mortice  in  the  post,  next  the  girder,  will 
be  2 feet  6J  inches  from  the  girder,  and  the  end  furthest  from  it  will 
be  3 feet.  The  bevel  of  braces  on  a regular  run  is  always  at  an  angle 
of  45°,  and  is  the  same  at  both  ends  of  the  brace. 

Pitch  of  the  Roof. 

In  this  building  the  roof  is  designed  to  have  a third  pitch ; that  is, 
the  peak  of  the  roof  would  be  one  third  the  width  of  the  building 
higher  than  the  top  of  the  plates,  provided  the  rafters  were  closely 
fitted  to  the  plates  at  their  outer  surfaces,  as  in  Plates  Nos.  3, 4,  and  6 ; 
but  it  is  common  in  barns,  and  sometimes  in  other  buildings,  as  has 
been  already  illustrated  in  Plate  5,  Fig.  3,  to  let  the  rafters  down  only 
half  their  width  upon  the  plates,  allowing  them  to  project  beyond  the 
plate,  so  that  in  this  case  the  peak  of  the  roof  is  10  feet  3 inches  above 
the  plates,  the  pitch  being  still  a third  pitch,  or  8 inches  rise  to  a foot 
run.  In  order  to  give  strength  to  the  mortices  for  the  upper  end 
girders,  these  girders  are  framed  into  the  corner  post  several  inches 
below  the  shoulders  of  the  post,  say  4 inches ; the  thickness  of  the 
plates  being  8 inches,  it  will  be  perceived  that  the  dotted  line,  AB, 
drawn  from  the  outer  and  upper  corner  of  one  plate  to  the  outer  and 
upper  corner  of  the  other,  is  just  1 foot  higher  than  the  upper  surface 
of  the  girder;  and  that  the  peak  of  the  roof  is  11  feet  3 inches  above 
this  girder.  The  length  and  bevels  of  the  rafters  can  be  found  as  al- 
ready described  in  Plate  3 and  Table  1. 

Purlins. 

The  purlin  plates  should  always  be  placed  under  the  middle  of  the 
rafters ; and  the  purlin  posts,  being  always  framed  square  with  the 
purlin  plates,  the  bevel  at  the  foot  of  these  posts  will  always  be  the 


BARN  FRAMING. 


57 


eame  as  the  upper  end  bevel  of  tbe  rafters  ;*  also,  the  bevel  at  each 
end  of  the  gable-end  girder  will  be  the  same,  since— the  two  girders 
being  parallel,  and  the  purlin  post  intersecting  them — the  alternate 
angles  are  equal.  (Prop.  XII.,  Cor.,  p.  29.)  The  length  of  the  gable- 
end  girder  will  be  equal  to  half  the  width  of  the  building,  less  18 
inches ; 6 inches  being  allowed  for  half  the  thickness  of  the  purlin 
posts,  and  6 inches  more  at  each  end  for  bringing  it  down  below  the 
shoulders  of  the  posts. 

Length  of  the  Purlin  Posts. 

In  order  to  obtain  the  length  of  the  purlin  posts,  let  the  learner  pay 
particular  attention  to  the  following  explanation  of  Fig.  2.  Let  the 
point  P represent  the  middle  point  of  the  rafter,  and  let  the  dotted 
line  PO  be  drawn  square  with  AB ; then  will  AG  be  the  J of  AB,  or 
7|  feet,  and  PC,  half  the  rise  of  the  roof,  will  be  5 feet,  and  PO  6 
feet.  The  purlin  post  being  square  with  the  rafter,  and  PO  being 
square  with  AB,  we  can  assume  that  PR  would  be  the  rafter  of  an- 
other roof  of  the  same  pitch  as  this  one,  provided  PO  were  half  its 
width,  and  OR  its  rise ; and  then,  since  we  know  the  length  of  PO, 
the  length  of  PR  could  also  be  found  by.  the  rafter  table  (No.  1,  Part 
IV.),  as  follows : — Width  of  building,  12  feet ; rise  of  rafter,  J of  12,  or 
4 feet ; hence,  length  of  rafter,  or  PR,  equals  7 feet  2 inches ; from 
this  deduct  half  the  width  of  the  rafter  and  the  thickness  of  the  pur- 
lin plate,  or  9 inches,  and  we  have,  6 feet  5t6$j  inches  as  the  length  of 
the  purlin  post,  from  the  shoulder  at  the  top  to  the  middle  of  the 
shoulder  at  the  foot.f  This  demonstration  determines  also  the  place 
of  the  purlin  post  mortice  in  the  girder;  for  AC  being  7|  feet,  and 
OR  being  4 feet,  bj  adding  these  together,  we  find  the  point  R,  the 


* This  fact  is  capable  of  a geometrical  demonstration;  for  the  triangle  FOR  is  similar 
to  the  triangle  ACP;  the  side  PR  in  one,  being  perpendicular  to  the  side  AP  in  the 
»tlier,  the  side  PO  being  also  perpendicular  to  AC,  and  the  side  RO  perpendicular  to 
PC.  (Part  I.,  Prop.  XXIX.)  Hence,  the  angles  opposite  the  perpendicular  sides  are 
equal ; and  we  have  angle  APC,  which  is  the  same  as  the  upper  end  bevel  of  the  rafter 
—being  parallel  with  it— equal  to  PRC,  the  angle  formed  by  the  purlin  post  and  th® 
girder  at  their  intersection  at  R. 

f The  following  geometrical  demonstration  of  the  above  proposition  is  subjoined.  In 
the  two  similar  triangles  ACP  and  POR,  the  sides  about  the  equal  angles  are  propor- 
tional (Def.  31);  and  we  have,  CP:  AC:  : OR:  OP;  but  CP  is  § of  AC;  consequently, 
OR  is  § of  OP.  But  OP  equals  6 feet;  hence,  OR  equals  4 feet.  Again,  the  triangle 
POR  being  right-angled  at  0,  then  POa-j-OR*=PR2.  4*=16,  and  6a=36 ; 3G- {-16=52, 
and  \/&2  P =7  ft  2.52  in.,  as  above. 


58 


CARPENTRY  MADE  EASY. 


middle  of  the  mortice,  to  be  11J  feet  from  the  outside  of  tbe  build- 
ing ; and  the  length  of  the  mortice  being  7J  inches,  the  distance  of  the 
end  of  the  mortice,  next  the  centre  of  the  building,  is  11  feet  9§  inches 
from  the  outside  of  the  building. 

Purlin  Post  Brace. 

The  brace  of  the  purlin  post  must  next  be  framed,  and  also  the 
mortices  for  it,  one  in  the  purlin  post  and  the  other  in  the  girder. 
The  length  of  the  brace  and  the  lower  end  bevel  of  it  will  be  the  same 
as  in  a regular  8 feet  run ; and  the  upper  end  bevel  would  also  be 
the  same,  provided  the  purlin  post  were  to  stand  perpendicular  to  the 
girder ; but,  being  square  with  the  rafter,  it  departs  further  and  fur- 
ther from  a perpendicular,  as  the  rafter  approaches  nearer  and  nearer 
toward  a perpendicular ; and  the  upper  end  bevel  of  the  brace  varies 
accordingly,  approaching  nearer  and  nearer  to  a right  angle  as  the 
bevel  at  the  foot  of  the  post,  or,  what  is  the  same  thing,  the  upper  end 
bevel  of  the  rafter  departs  further  and  further  from  a right  angle. 
Hence,  the  hevel  at  the  top  of  this  brace  is  a COMPOUND  BEVEL,  found  by 
adding  the  lower  end  bevel  of  the  brace  to  the  upper  end  bevel  of  the  rafter  * 
(See  Plate  8.) 


Purlin  Post  Brace  Mortices. 

In  framing  the  mortices  for  the  purlin  post  braces,  it  is  to  be  ob- 
served, also,  that  if  the  purlin  post  were  perpendicular  to  the  girder, 
the  mortices  would  each  of  them  be  8 feet  from  the  heel  of  the  post ; 
but  as  the  post  always  stands  back,  so  the  distance  will  always  be 
more  than  3 feet  from  the  heel  of  the  post;  and  the  sharper  the  pitch 
of  the  roof,  the  greater  this  distance  will  be.  Hence  the  true  distance 
on  the  girder  for  the  purlin  post  brace  mortice  is  found  by  adding  to 
3 feet  the  rise  of  the  roof  in  running  3 feet;  which,  in  this  pitch  of  8 
inches  to  the  foot,  is  2 feet  more,  making  5 feet,  the  true  distance 
of  the  furthest  end  of  the  mortice  from  the  heel  of  the  purlin  post. 

The  place  in  the  purlin  post  for  the  mortice  for  the  upper  end  of 
the  brace  may  be  found  from  the  rafter  table,  by  assuming  that  Ras 


• This  proposition  is  capable  of  demonstration,  thus:  The  angle  PxM  equals  the  sum 
f the  angles  MRz  and  xMR,  since  PxM  is  the  exterior  angle  of  the  triangle  MRx, 
formed  by  producing  the  base  Rx  in  the  direction  RxP.  (See  Prop.  XVI.,  Cor.)  But 
the  angle  PxM  is  the  upper  end  bevel  of  the  purlin  post  brace;  therefore,  it  is  equal  to 
the  Bum  of  the  two  bevels,  one  at  the  foot  of  the  brace  and  the  other  at  the  foot  of  tho 
post,  as  above 


BARN  FRAMING. 


59 


would  be  the  rafter  of  another  roof  of  the  same  pitch  as  this  one,  if 
xy  were  half  the  width,  and  ?/R  the  rise.  For  then,  since  xy  equals 
3 feet,  we  should  have  width  of  building  equal  6 feet,  rise  of  rafter, 
one  third  pitch,  gives  ?/R  equal  2 feet ; and  hence  xR  would  equal 
3 feet  7.26  inches,  the  true  distance  of  the  upper  end  of  the  mortice 
from  the  heel  of  the  purlin  post.* 


* The  same  proposition  is  demonstrated  by  Geometry,  as  follows ; xy  being  parallel 
with  PO,  the  two  triangles  RPO  and  R xy  are  similar,  (Geom.,  Prop.  XXIX),  hence  the  sides 
opposite  the  equal  angles  are  proportional,  and  we  have  Rx  ; RP  • : xy  j PO.  But  we 
have  already  found  PO  to  equal  6 feet,  and  xy  equal  to  3 feet,  and  RP  equal  to  7 feet 
H.62  inches.  Hence, 

6 s 3 : : 7 ft.  2.52  in.  s 3 ft  7.26  in.  Answer  as  above. 


PLATE  8. 


UPPER  END  BEVEL  OF  PURLIN  POST  BRACES. 

Plate  8 is  designed  to  illustrate  the  manner  of  finding  the  upper 
end  bevel  of  purlin  post  braces,  to  which  reference  is  made  from  the 
preceding  Plate. 

In  Fig.  1,  let  AB  represent  the  extreme  length  of  the  brace  from 
toe  to  toe,  the  bevel  at  the  foot  having  been  already  cut  at  the  proper 
angle  of  45  degrees.  Draw  BO  at  the  top  of  the  brace,  at  the  same 
bevel ; then  set  a bevel  square  to  the  bevel  of  the  upper  end  of  the 
rafter,  and  add  that  bevel  to  BC;  by  placing  the  handle  of  the  square 
upon  BC  and  drawing  BD  on  the  tongue.  This  is  the  bevel  required. 

Fig.  2 shows  another  method  of  obtaining  the  same  bevel.  Let  the 
line  AB  represent  the  bevel  at  the  foot  of  the  brace,  drawn  at  an  angle 
of  45  degrees.  Draw  BD  at  right  angles  with  AB,  and  draw  BO  per- 
pendicular to  AD,  making  two  right-angled  triangles.  Then  divide 
the  base  of  the  inner  one  of  these  triangles  into  12  equal  parts,  for 
the  rise  of  the  roof.  Then  place  the  bevel  square  upon  the  bevel  AB, 
at  B,  and  set  it  to  the  figure  on  the  line  CD,  which  corresponds  with 
the  pitch  of  the  roof.  This  will  set  the  square  to  the  bevel  required 
for  the  top  of  the  brace.  In  this  figure  the  bevel  is  not  marked  upon 
the  brace,  but  the  square  is  properly  set  for  a pitch  of  8 inches  to  the 
foot,  or  a one  third  pitch.  The  square  can  now  be  placed  upon  the 
top  of  the  brace,  and  the  bevel  marked, 

(60) 


Plate  8. 


♦ 


Plate  9 


Scale  S feet  to  the  inch. 


Plate  10. 


B 


L_!™ 


PLATES  9 & 10. 


Plates  9 and  10  exhibit  the  side  and  end  elevations  of  a building 
designed  for  a warehouse,  or  mill. 

Length  of  building,  50  feet ; 

Width  of  building,  40  feet; 

Height  of  building  from  the  foundation  to  the  top  of  plate?,  88  feet ; 

Main  timbers,  12  inches  square; 

Door  posts,  10  by  12 ; 

Purlin  posts,  8 by  10 ; 

Plates  and  purlin  plates,  8 by  8 ; 

Braces,  4 by  6 ; 

Lower  joists,  8 by  12 ; 

Upper  joists,  8 by  10; 

Studding,  2 by  8 ; 

Rafters,  2 by  6. 

The  posts  are  framed  in  sections,  one  story  at  a time,  on  account 
of  the  difficulty  in  procuring  long  timbers,  also  for  facility  in  raising 
the  building ; for,  by  this  means,  each  story  can  be  raised  separately. 
It  has  also  been  proved  by  experience,  that  when  the  timbers  are 
locked  together  as  represented  in  the  Plate,  this  mode  of  building  is 
equally  strong  as  to  have  the  posts  in  one  length.  The  ends  of  the 
joists  are  sized  to  a uniform  width,  and  placed  upon  the  timbers, 
the  crowning  side  up  ; the  studs  are  morticed  into  the  timbers  as  usual. 
The  roof  is  framed  to  a quarter  pitch,  and  the  braces  to  a regular 
8 feet  run.  Plate  3 describes  the  manner  of  obtaining  the  bevels  of 
the  rafters  and  gable-end  studding.  Plates  7 and  8 show  the  manner 
of  obtaining  the  bevels  of  the  purlin  posts  and  braces.  Plate  4 gives 
the  method  of  finding  the  length  of  the  gable-end  studding. 

Cripple  Studs. 

The  length  of  the  cripple  studs,  which  are  to  be  nailed  to  the  braces, 
depends  upon  the  run  of  the  braces.  The  braces  in  this  building,  being 
on  a regular  run,  are  all  set  at  an  angle  of  45  degrees,  so  the  bevel  of 
the  cripple  studs  will  be  the  same;  and  the  rise  of  the  brace  being  equal 
to  the  run.  the  length  of  each  cripple  stud  will  be  equal  to  the  height 
of  the  post  from  the  sill  to  the  toe  of  the  brace,  added  to  the  distance 

(fill 


62 


CARPENTRY  MARE  EASY* 


of  the  stud  from  the  post.  In  this  building,  the  height  of  the  brace 
from  the  sill  to  the  toe  of  the  brace  in  the  first  story  is  8 feet,  and  the 
first  stud  being  14  inches  from  the  post,  and  2 inches  thick,  the  length 
of  the  first  cripple  stud  will  be  16  inches  longer  than  the  height  of 
the  post  from  the  sill  to  the  toe  of  the  brace,  or  9 feet  4 inches ; and 
the  length  of  the  next  cripple  stud  will  be  16  inches  more,  or  10  feet 
8 inches. 

It  now  remains  to  determine  the  bevels  and  the  lengths  of  those 
cripple  studs  in  the  gable  end,  which  are  to  come  against  the  purlin 
posts.  Having  already  (Plate  7)  found  the  bevel  at  the  foot  of  the 
purlin  post  equal  to  the  upper  end  bevel  of  the  rafters,  it  will  follow 
that  the  bevel  of  the  cripple  studs  upon  the  purlin  post  is  equal  to  the 
lower  end  bevel  of  the  rafters*  The  length  of  the  cripple  studs  stand- 
ing between  the  rafter  and  the  purlin  posts  depends  both  upon  the  rise 
of  the  roof  and  the  rise  of  the  purlin  post;  but  the  purlin  post 
being  set  square  with  the  rafter,  its  rise  is  always  the  same  as  the  run 
of  the  rafter,  and  its  run  is  the  same  as  the  rise  of  the  rafter. 

Hence,  for  finding  the  length  of  a cripple  stud,  standing  in  any 
building  between  the  rafter  and  the  purlin  post,  at  a certain  horizontal 
distance  from  the  top  of  the  purlin  plate,  we  have  the  following  Rule: 
A dd  the  RISE  of  the  roof  in  RUNNING  the  given  distance  to  the  RUN  of  the  roof 
in  RISING  the  given  distance  ; the  sum  will  give  the  length  of  the  cripple  stud. 

For  example,  in  this  plate,  suppose  the  cripple  stud  / to  be  18  inches 
from  the  top  of  the  purlin  plate,  horizontal  distance,  then  the  rise  of 
the  roof  on  a quarter  pitch  in  running  18  inches  would  be  9 inches, 
and  the  run  of  the  roof  in  rising  18  inches  would  be  36  inches;  so  that 
the  length  of  / is  45  inches.  The  stud  marked  M being  16  inches 
from  /,  the  additional  rise  is  8 inches,  and  the  additional  run  is  32 
inches,  so  that  M is  40  inches  longer  than  /. 

Note  on  Bevels.—  The  bevels  in  a frame  of  this  kind  are  only  four  in  number:— 

1.  The  hevel  of  the  upper  end  of  the  rafter. 

2.  The  bevel  of  the  foot  of  a rafter. 

8.  The  bevel  of  the  braces,  &c — equal  to  45  degrees. 

4.  The  bevel  of  the  upper  end  of  the  purlin  post  brace,  always  equal  to  the  sum  of 
the  first  and  third.  Balloon  frames  have  but  two  bevels — the  first  and  second  above 
mentioned. 


• Demonstrated  as  follows.  The  triangle  ABC  is  similar  to  the  triangle  DEF,  since 
the  sides  uf  the  one  are  perpendicular  to  the  sides  of  the  other;  consequently  the  angle* 
opposite  the  perpendicular  sides  are  equal.  (Geom.,  Prop.  29.) 

The  side  FE,  in  one  triangle,  is  perpendicular  to  the  side  BC  in  the  other;  hence,  the 
angle  A=  angle  D.  The  angle  A is  the  lower  bevel  of  the  rafters,  and  the  angle  D is  the 
bevel  ot  tl  e cripple  stud  on  the  purlin  post. 


- 


V 


Plate  11. 


PLATE  11. 


Plate  11  is  designed  to  represent  two  modes  of  framing  braces  in  a 
self-supporting  or  trussed  partition.  Where  the  span  is  considerable,  there 
being  no  support  beneath  except  the  exterior  wall,  some  mode  of 
bracing  is  indispensable.  These  plans  are  exhibited  as  being  practi- 
cable and  secure. 

The  first  plan  gives  opportunities  for  two  or  even  three  openings. 

The  second  plan  will  be  most  convenient  where  only  one  opening 
is  desired. 

The  size  of  these  brace  timbers  should  be  in  proportion  to  the  width 
of  the  building,  and  the  weight  which  the  partition  is  to  sustain.  If 
they  are  ten  or  twelve  inches  square,  they  will  safely  sustain  a brick 
wall  built  upon  the  partition. 

Fig.  8 is  designed  to  show  the  proper  mode  of  trussing  a beam  over 
a barn  floor,  or  in  frpnt  of  a church  gallery,  or  any  other  situation 
where  it  is  inconvenient  to  support  it  by  posts. 


(«3) 


PLATE  12. 


SCARFING. 

T/>vs  Plate  exhibits  several  designs  for  scarfing  or  splicing  timber. 
The  length  of  the  splice  should  be  about  four  times  the  thickness  of 
the  timber ; and  when  the  joint  is  beveling,  it  will  be  found  the  best 
and  most  expeditious  way,  first,  to  prepare  an  exact  pattern  of  boards, 
and  then  to  frame  the  timbers  by  the  pattern : by  this  means  a perfect 
ioint  can  be  made. 

Straps  and  Bolts. 

Fig.  4 is  spliced  by  strapping  pieces  of  plank  upon  the  upper  and 
lower  sides  of  the  joint,  and  securing  them  with  bolts  of  £ inch  or  1 
inch  in  diameter,  according  to  the  size  of  the  timber. 

Figs.  5,  7,  and  8,  have  iron  straps  bolted  in  a similar  manner. 

Fig  9 exhibits  a strong  mode  of  splicing  timbers  where  they  are 
doublsd  throughout  their  whole  length,  for  a very  long  span,  such  as 
roofs  in  churches. 

Those  styles  which  are  numbered  1,  2,  7,  and  5,  are  recommended 
as  being  the  best  in  proportion  to  the  cost. 

(64) 


Plate  13 . 


PLATE  13. 


Scarfing,  whenever  it  is  practicable,  should  be  made  directly  over  a 
post,  when  a simple  and  inexpensive  style,  such  as  is  exhibited  in  Plate 
13,  will  be  found  sufficiently  strong.  Indeed,  it  is  hardly  possible 
to  find  a stronger  mode  of  scarfing  than  that  illustrated  in  Fig.  1 ; and 
yet,  being  supported  by  the  post,  the  design  is  more  simple  than  most 
of  those  represented  in  Plate  12. 

In  this  design,  the  head  of  the  post  is  framed  into  a bolster,  which 
ought  to  be  fully  equal  in  size  to  the  timber  which  it  is  required  to 
support ; in  heavy  frames  it  should  be  about  6 feet  long,  the  braces 
being  framed  on  a 4 feet  run.  The  bolster  is  secured  to  the  timber 
with  inch  bolts,  as  represented  in  the  figure. 

Fig.  1 also  illustrates  the  proper  mode  of  framing  braces ; the  dotted 
lines  show  the  form  of  the  tenons,  and  the  notchings  in  the  post  and 
the  girder  represent  the  facing  of  the  mortices.  These  are  usually 
notched-in  half  an  inch  at  least,  in  order  to  give  all  possible  support 
to  the  toe  of  the  brace ; and  the  measurement,  for  the  length  and  the 
run  of  the  braces,  must  be  from  the  furthest  point  of  the  face  of  the 
mortice,  inside  of  the  notch,  and  not  from  the  outside  of  the  rough 
timber. 

The  shoulder  of  the  post  is  also  notched  or  sized  into  the  bolster  an 
inch,  and  the  bolster  is  locked  into  the  girder  2 inches ; and,  if  the 
timber  is  12  inches  square,  the  shoulder  of  the  post  is  10  inches  lower 
than  it  would  be  without  any  bolster ; and  the  carpenter  must,  of 
course,  make  his  brace  mortices  in  the  post,  accordingly,  10  inches 
nearer  the  shoulder  than  usual. 

Figs.  2 and  3 are  designed  to  represent  a less  expensive  mode  of 
scarfing  on  posts.  In  these  plans  the  tenons  extend  quite  through, 
and  are  double-pinned  to  both  timbers,  as  represented  in  the  plate. 
This  mode  is  sufficiently  strong  for  scarfing  plates  and  purlin  plates. 

5 " (65) 


» 


PLATE  14. 


FLOORS  IN  A BRICK  BUILDING. 

Plate  14,  Fig.  1,  exhibits  the  ground  plan  of  one  room,  18  feet  wila 
The  joists  are  2 by  10,  18  feet  long. 

Trimmer  Joists. 

Those  marked  A,  B,  and  C,  at  each  side  of  the  fire-place,  are  billed 
trimmer  joists ; they  are  4 by  10 ; or  they  may  be  made  by  spiking  two 
common  joists  together,  as  represented  in  the  plate. 

A course  of  bridging  is  represented  at  D,  as  described  in  Plate  6, 

Fig.  2 and  Fig.  5 exhibit  the  method  of  framing  the  tenons  of  the 
common  joists. 

Fig.  3 shows  the  mode  of  framing  together  the  trimmer  joists  at  the 
corners  of  the  hearth 

Fig.  4 shows  the  beveled  ends  of  the  joists  where  they  are  set  into 
the  brick  wall.  They  are  beveled  in  the  manner  represented,  that 
the  springing  of  the  joists  may  not  endanger  the  wall ; and,  in  case  of 
fire,  the  joists  may  burn  and  fall  out  without  destroying  the  wall 
(66) 


Plate  15. 


PLATE  15. 


CIRCULAR  CENTRES. 

Plate  15  exhibits  several  designs,  more  or  less  convex,  for  the  con 
struction  of  centres,  which  are  skeletons  used  by  stone  and  brick 
masons  to  build  an  arch  upon,  but  which  are  to  be  taken  away  when 
the  arch  is  sprung  and  the  mortar  set. 

First,  draw  the  line  AB  upon  a floor,  two  inches  less  than  the  width 
of  the  skeleton  required,  and  take  OA  as  a radius,  and  describe  the 
semi-circumference.  Inch  boards  are  then  to  be  fitted  to  this  curved 
line,  and  their  ends  beveled  to  fit  each  other,  as  represented  in  the 
Plate.  The  bevel  is  determined  by  simply  drawing  a straight  line  from 
any  point  of  the  curve  to  the  central  point  0.  Other  boards  are  then 
to  be  nailed  over  the  joints,  on  the  inside  of  these ; and  the  long  brace 
AB  nailed  at  each  end  at  the  bottom. 

Having  prepared  the  other  end  of  the  skeleton  in  the  same  manner, 
strips  of  boards,  an  inch  thick,  two  inches  wide,  and  of  a length  equal 
to  the  thickness  of  the  wall,  are  nailed  upon  their  convex  edges,  as 
represented. 

Should  the  arch  have  more  than  12  feet  span,  it  would  be  proper 
to  use  thicker  boards ; but  for  any  thing  less  than  12  feet,  inch  boards 
are  amply  sufficient 


PLATE  16. 

ELLIPTICAL  CENTRES. 

This  Plate  illustrates  the  manner  of  constructing  elliptical  centres. 
The  elliptical  curve  is  described  most  accurately  by  means  of  a tram - 
mel,  the  construction  and  use  of  which  are  explained  in  Plate  2,  page  20. 

In  order  to  describe  the  curve  for  these  centres,  take  AB  equal  to 
the  span  of  the  arch,  less  2 inches,  and  set  the  trammel  so  that  the 
intersection  of  the  arms  will  fall  upon  O,  the  middle  point  of  AB. 
Then  set  the  pin  B,  so  that  PB  will  equal  the  height  of  the  arch  less 
1 inch ; and  set  the  pin  C so  that  PC  will  equal  AO. 

m 


Tlu-oJ.^ou'jArtU  S;  Son  i ini  ’ 


\ 


* 


PLATE  17. 


ARCHES. 

It  is  also  tlie  business  of  the  carpenter  to  prepare  patterns  for  stone- 
cutters, by  which  they  are  to  cut  their  stones  to  fit  arches  of  any  de- 
sired form.  This  Plate  exhibits  seven  different  styles  of  arches,  with 
the  most  accurate  and  convenient  modes  of  drawing  them,  and  of 
dividing  them  into  proper  sections  or  patterns  for  the  arch-stones. 

! 

Fig.  1 represents  an  elliptical  arch,  drawn  by  means  of  a trammel* 
as  has  been  already  described.  The  arch  is  divided  into  blocks  of 
proper  form  and  size,  by  first  dividing  the  curve  into  any  desirec) 
number  of  equal  spaces ; then,  wherever  a joint  is  required,  first  drkw 
a tangent  to  the  curve  at  that  point,  and  then  a line  perpendicular  td 
this  tangent  will  divide  the  arch  properly. 

Fig.  2 exhibits  the  Tudor  arch,  drawn  in  two  rampant  semi-elliptical 
curves,  as  represented  in  the  Plate.  The  same  rule  is  to  be  observed 
as  before  for  finding  the  joints  of  this  arch,  and  for  dividing  it  cor- 
rectly into  proper  patterns. 

Fig.  3 is  a semicircular  arch.  This  is  most  easily  and  correctly 
jointed  by  drawing  a radius  from  the  centre  0 to  any  point  of  the 
curve  where  a joint  is  desired. 

Fig.  4 is  a Gothic  arch,  described  by  two  equal  radii  from  the 
points  0,  O,  as  centres,  and  jointed  from  the  same  points. 

Fig.  5 is  so  similar  to  the  last,  as  to  require  no  further  description. 

Figs.  6 and  7 exhibit  two  depressed  segmental  arches  of  the  same 
span,  but  representing  different  degrees  of  curvature,  one  being  drawn 
with  a longer  radius  than  the  other,  and  both  jointed  by  radii  drawn 
from  the  common  centre  0. 


(69) 


PLATE  18. 

♦ HIP  ROOFS. 

Plate  18,  Fig.  1,  exhibits  the  plan  of  a hip  roof  in  a building  50  feet 
long  and  40  feet  wide,  the  rise  of  the  roof  being  5 inches  to  the  foot. 
AB,  CD,  and  EF,  are  the  upper  girders,  which  are  trussed  for  support- 
ing the  roof,  with  short  principal  rafters  and  straining  beams,  as  rep- 
resented in  the  Plate  at  CD. 

GrH,  IH,  KL,  and  ML,  are  the  hip  rafters,  and  HL  the  ridge  pole. 
The  purlin  plates  are  placed  upon  the  principal  rafters,  and  the  strain- 
ing beams  are  framed  on  a level  with  the  purlin  plates,  so  that  the 
end  ones  may  answer  the  double  purpose  of  straining  beams  and 
purlin  plates  also. 

The  lengths*  and  bevels  of  the  common  rafters,  in  the  middle  of  the 
building,  the  upper  ends  of  which  rest  against  the  ridge  pole,  are 
found  as  usual  in  common  roofs.  (Table  No.  1.) 

Hip  Rafters. 

The  length  of  the  hip  rafters  is  given  in  the  Hip  Rafter  Table,  (No. 
2,  p.  133)  where  the  rule  for  obtaining  it  is  fully  explained. 

In  addition  to  the  two  bevels  common  to  all  rafters,  namely,  the 
upper  end  bevel  and  the  lower  end  bevel,  hip  rafters  have  two  other 
bevels,  which  are  the  side  bevel , as  it  is  called,  being  the  angle  which 
the  two  hip  rafters  make  with  each  other  at  their  intersection,  and  the 
bacJcing}  which  is  the  angle  made  by  the  intersection  of  the  side  roof 
with  the  end  roof. 

Side  Bevel  of  the  Hip  Rafters, 

It  is  evident  that  if  there  were  no  pitch  to  the  roof,  this  bevel  would 
be  45  degrees,  since  the  two  hip  rafters  would  be  perfectly  square  with 
each  other ; but  as  soon  as  the  roof  begins  to  rise,  the  hip  rafters  are 
no  lunger  square  with  each  other,  but  begin  to  approach  a parallel. 


• Throughout  this  work,  the  length  of  rafters  is  computed  from  the  upper  and  outer 
corners  of  the  plates  to  the  very  peak  of  the  roof,  without  allowances  for  the  projection 
of  the  rafters  or  the  thickness  of  the  ridge  poles.  In  case  the  building  has  a ridge  pole, 
therefore,  it  will  be  necessary  to  deduct  one  half  the  thickness  of  it  from  the  length  as 
given,  measuring  for  this  deduction  square  from  the  down  bevel,  and  not  lengthwise  of 
the  rafter. 

rroj 


HIP  ROOFS. 


71 


till,  in  a very  steep  roof,  as  tliat  of  a tower  or  steeple,  they  are  nearer 
parallel  than  square  with  each  other.  The  side  bevel,  therefore,  La 
always  greater  than  45  degrees,  and  is  obtained  from  the  square  by 
taking  the  length  of  the  hip  rafter  on  the  blade,  and  its  run  on  the 
tongue;  the  blade  then  shows  the  side  bevel  of  the  hip  rafter  re- 
quired. 

Down  Bevel  of  the  Hip  Rafters. 

The  upper  end  bevel  is  commonly  called,  in  hip  rafters,  the  down 
bevel.  It  is  always  square  with  the  lower  end  bevel,  the  one  being 
the  complement  of  the  other.  This  bevel  varies  with  the  pitch  of  the 
roof;  for  the  pitch  of  the  hip  rafter  always  has  the  same  propor- 
tion to  the  pitch  of  the  common  rafter,  that  the  run  of  the  common 
raftci  has  to  the  run  of  the  hip  rafter,  or  that  the  side  of  a square  has 
to  ns  diagonal ; for,  if  we  let  0 represent  the  foot  of  the  perpendicular 
let  fall  from  L upon  the  middle  of  the  girder,  CD,  then  ODPM  is  a 
square,  of  which  OD,  the  run  of  a common  rafter,  is  the  side,  and  OM, 
the  run  of  the  hip  rafter,  is  the  diagonal.  Now,  it  is  a well-known 
principle  in  mathematics,  that  the  side  of  any  square  is  proportioned  to 
its  diagonal , as  ONE  is  to  the  SQUARE  ROOT  OF  TWO,  (Prop.  XXIV., 
Cor.);  or,  as  1 is  to  1.4142;  or,  as  12  inches  are  to  17  inches  nearly. 
When  the  common  rafter,  therefore,  has  5 inches  rise  to  12  inches  run, 
the  hip  rafter  of  the  same  roof  has  5 inches  rise  to  17  inches  run ; and 
when  the  common  rafter  has  6 inches  rise  to  12  inches  run,  the  hip 
rafter  has  6 inches  rise  to  17  inches  run;  &c. 

Prom  these  demonstrations  we  derive  the  following  Rule  for  find- 
ing the  down  bevel  and  the  lower  end  bevel  of  the  hip  rafters.  Take 
17  inches  on  the  blade  of  the  square  for  the  run , and  the  rise  of  the  roof  to 
the  foot  on  the  tongue.  The  tongue  will  give  the  down  bevel,  or  upper 
end  bevel,  and  the  blade  the  lower  end  bevel. 

Backing  of  the  Hip  Rafters. 

This  is  found  on  the  square  by  taking  the  length  of  the  hip  rafter 
on  the  blade,  and  the  rise  of  the  roof  on  the  tongue ; the  bevel  of  the 
tongue  will  be  the  backing  required.  For  illustration,  in  this  build- 
ing the  length  of  the  hip  rafters,  as  given  in  the  Table,  equals  29  J 
feet  nearly , and  the  rise  of  the  roof  is  8 J feet.  Take  proportional  parts 
of  each  on  the  square,  and  the  tongue  will  give  the  backing;  that  is, 
place  the  square  upon  a strait  edge,  with  the  blade  at  the  14f  inch 
mark,  and  the  tongue  at  tne  4J  inch  mark,  and  draw  a scratch  along 
the  tongue;  then  set  a bevel  square  to  the  angle  which  this  scratch 
makes  with  the  straight  edge,  and  it  is  the  backing  required. 

For  further  illustration  see  Plate  39,  and  page  111. 


72 


CARPENTRY  MADE  EASY. 


Lengths  and  Bevels  of  the  Jack  Rafters. 

Since  the  pitch  of  the  jack  rafters  is  the  same  as  that  of  the  common 
rafters,  the  longest  jack  rafter,  the  upper  end  of  which  rests  against 
the  end  of  the  ridge  pole,  is  of  the  same  length  as  the  common  rafters, 
as  given  in  the  Common  Rafter  Table,  less  half  the  thickness  of  the 
hip  rafters  at  their  side  bevel  on  the  ridge  pole.  The  difference  m 
length  between  the  longest  jack  rafter  and  the  next  one,  or  between 
any  two  adjacent  ones,  is  equal  to  their  distance  apart,  added  to  the 
gain  of  the  rafter  in  running  that  distance. 

For  example,  in  this  building,  the  width  being  40  feet,  the  run  of 
the  jack  rafter  is  20  feet,  or  240  inches ; and  its  lengthy  on  a pitch  of  5 
inches  rise  to  the  foot,  is  260  inches;  therefore,  its  gain  is  20  inches  in 
running  20  feet,  or  an  inch  to  a foot.  The  jack  rafters  being  2 feet 
apart,  the  difference  in  length  between  any  two  adjacent  ones  is,  there- 
fore, 2 feet  2 inches. 

Or,  the  length  of  the  shorter  jack  rafters  may  be  obtained  from  that 
of  the  longest  one,  by  dividing  the  length  of  the  longest  one  by  the 
number  of  spaces  between  the  longest  one  and  the  corner  of  the  build- 
ing ; the  quotient  will  be  the  length  of  the  shortest  one,  and  it  will  be 
also  the  difference  between  any  two  adjacent  ones. 

The  down  bevel  and  the  lower  end  bevel  are  the  same  as  the  upper  and 
lower  end  bevels  of  the  common  rafters. 

The  side  bevel  of  the  jack  rafters  is  always  more  than  45°,  for  a simi- 
lar reason  as  that  given  in  the  description  of  the  side  bevel  of  the  hip 
rafters ; and,  since  all  the  jack  rafters  have  the  same  pitch  as  the  com- 
mon rafters,  we  have  the  following 

Buie  for  obtaining  the  side  bevel  of  the  jack  rafters. — Take  the  length  of 
a common  rafter  on  the  blade  of  a square,  and  its  run  on  the  tongue— 
or  proportional  parts  of  each.  The  bevel  on  the  blade  is  the  side  bevel 
of  all  the  jack  rafters  in  the  frame. 

Or,  take  the  length  of  the  longest  jack  rafter  on  the  blade,  and  half 
the  width  of  the  building  on  the  tongue,  or  proportional  parts  of  each, 
aud  the  bevel  on  the  blade  will  be  the  required  side  bevel. 

Remark. — In  a hip  roof  which  is  perfectly  square,  the  hip  rafters 
need  have  no  side  bevel ; for  two  of  them  can  be  cut  of  full  length, 
and  set  up,  opposite  each  other  first,  with  their  down  bevels  resting 
full  against  each  other  like  common  rafters;  the  other  two  hip  rafters 
can  then  be  set  against  these,  having  been  cut  off  haif  their  thickness 
shorter  than  the  full  length, 


Plate 19 


i 

‘ 


Sca  le  /Oft.  to  the  r/reh . 


HIPS  AND  VALLEYS. 

Plate  19,  Fig.  1,  represents  tlie  liip  roof  of  a building  consisting  oi 
& main  portion  and  a wing,  the  walls  of  the  wing  being  of  the  same 
height  as  those  of  the  main  building.  The  main  building  being  24 
by  30  feet,  and  the  wing  10  by  20  feet,  the  roof  of  the  wing  being  of 
the  same  pitch,  will  not  rise  as  high  as  that  of  the  other  part. 

The  timbers  AB  and  CD  at  the  intersection  of  these  roofs  are  called 
valley  rafters.  The  upper  end  of  the  first  one,  AB,  is  extended  to  the 
ridge  pole  of  the  main  building ; the  other  valley  rafter  is  supported 
by  the  first  one,  by  being  spiked  to  it  at  their  intersection  at  C.  From 
that  point  of  intersection,  a ridge  pole  extends  to  the  intersection  of 
the  end  hip  rafters  at  E.  This  ridge  pole  is  equal  in  length  to  the 
whole  width  of  the  wing ; the  valley  rafter  and  hip  rafter  on  each  side 
being  parallel  with  each  other. 

The  lengths  and  bevels  of  the  various  rafters  are  found  as  explained 
in  the  preceding  Plate. 

TRAPEZOIDAL  HIP  ROOFS. 

Fig.  2 exhibits  the  plan  of  a hip  roof  for  a building  constructed  in 
the  form  of  an  irregular  square,  or  trapezoid  ; the  side  CD  being  4 feet 
longer  than  the  side  AB,  the  width  24  feet  and  the  rise  5 inches  to  the 
foot. 

The  length  and  bevels  of  the  common  rafters  and  of  the  hip  rafters 
on  the  square  end  of  the  frame  are  obtained  in  the  same  manner  as  de- 
scribed, in  the  regular  hip  roof.  The  lengths  and  bevels  of  the  two 
hip  rafters  BF  and  DF  on  the  beveled  end  of  the  frame  are  unlike  each 
other,  and  unlike  those  on  the  square  end,  one  being  longer  and  the 
other  shorter  than  those.  As  such  buildings  are  comparatively  rare,  it 
has  not  been  deemed  necessary  to  encumber  this  work  with  a table 
for  such  rafters,  but  the  facts  and  principles  applicable  to  such  frames 
are  here  stated. 

Lengths  of  the  Irregular  Hip  Rafters. 

If  this  end  of  the  building  were  square,  then  BGr  and  DI  would 
«ach  of  them  be  equal  to  half  the  width  of  the  building,  or  12  fees,* 

(78) 


74 


CARPENTRY  MADE  EASY. 


but  since  one  side  of  the  building  is  4 feet  shorter  than  the  other  side, 
the  short  side  is  2 feet  less  than  12,  and  the  long  side  2 feet  more  than 
12  ; or  BG  equals  10  feet,  and  DI  equals  14  feet. 

The  length  of  the  common  rafters  GF  and  IF  is  found  by  the  table 
to  equal  13  feet ; and  the  two  triangles  GFB  and  IFD  being  right 
angled,  we  are  now  furnished  with  the  means  of  finding  the  lengths  of 
the  two  hip  rafters  BF  and  DF,  by  the  use  of  that  familiar  principle  in 
mathematics,  that  in  every  right-angled  triangle  the  square  of  the  hypote- 
nuse is  equal  to  the  sum  of  the  squares  of  the  other  two  sides.  (Part  I.,  Prop. 
XXIV.)  For  in  the  right-angled  triangle  GFB,  we  have  the  side  GB 
equal  to  10  feet,  which  we  reduce  to  inches,  to  secure  greater  accuracy  in 
calculation.  GB  then  equals  120  inches,  and  GF,  a common  rafter, 
equals  13  feet  or  156  inches.  The  square  of  120  is  14,400,  and  the 
square  of  156  is  24,836;  the  sum  of  these  squares  is  38,786,  which  is 
the  same  as  the  square  of  the  hypotenuse  BF ; and  by  extracting  the 
square  root  of  38,736  inches,  we  have  195  inches  and  81  hundredths  of 
an  inch,  or  16  feet  4.81  inches,  as  the  length  of  the  hip  rafter  BF. 

The  length  of  DF  is  found  in  a similar  manner  from  the  triangle  IFD, 
DI  being  14  feet  or  168  inches,  the  square  of  which  is  28,224,  to  which 
add  the  square  of  IF,  equal  to  that  of  GF,  found  above  to  be  24,336, 
The  sum  of  these  squares  is  52,560  inches,  of  which  the  square  root  is 
229.25  inches,  or  19  feet  1.25  inches,  the  true  length  of  the  longest  hip 
rafter. 

Bevels  of  the  Irregular  Hip  Kafters. 

The  down  level  and  the  lower  end  level  of  these  irregular  hip 
rafters,  like  those  of  all  other  rafters,  are  square  with  each  other : and 
are  found  together  on  the  square  by  taking  the  run  on  the  blade  and 
the  rise  on  the  tongue  ; the  tongue  will  give  the  down  bevel,  and  the 
blade  the  lower  end  bevel. 

Note. — Take  the  square  of  BG,  and  the  square  of  DI,  respectively;  add  to  each  the 
a^uaj-o  of  half  the  width  of  the  building;  extract  the  square  root  of  each  of  these  sums, 
and  they  will  give  the  runs  of  BF  and  DF,  respectively. 

Th  is,  the  square  of  BG,  10  feet,  or  120  inches,  is  14,400  inches;  the  square  of  DI,  34 
f«et,  or  168  inches,  is  28,224  inches;  and  the  square  of  half  the  width  of  the  building, 
12  feet,  or  144  inches,  is  20,736  inches;  this  added  to  14,400  is  36,136  inches,  the  square 
root  of  which  is  187.44  inches,  or  15  feet  7.44  inches,  which  is  the  run  of  BF. 

Again,  20,736  added  to  28,224  equals  48,960,  the  square  root  of  which  is  221.26  inches, 
or  18  feet  5.26  inches,  the  run  of  DF. 

Thesidelevel  of  irregular  hip  rafters  is  obtained  by  adding  together  the 
distance  from  the  foot  of  the  hip  rafter  to  the  foot  of  the  first  common 


TRAPEZOIDAL  HIP  ROOFS. 


75 


rafter,  and  the  gain  of  the  hip  rafter ; then  take  this  sum  on  the  tonguo 
of  a square,  and  half  the  width  of  the  building  on  the  blade ; and  the 
tongue  will  give  the  side  bevel  required. 

Backing  of  Hip  Eafters  on  Trapezoidal  and  other  Irregular  Roofs, 

As  the  oblique  angles  of  trapezoidal  and  other  irregular  buildings 
are  liable  to  many  variations,  so  the  backing  of  the  hip  rafters  must 
also  vary  on  different  roofs.  The  simplest  and  most  practical  manner 
of  finding  these  backings,  when  thus  irregular,  is  to  take  a small  square 
block,  and  bevel  one  end  of  it  to  the  same  bevel  as  the  lower  end  of 
the  hip  rafter  in  question.  Then  place  this  beveled  end  upon  the 
plates,  just  as  the  hip  rafter  is  to  be  placed,  at  the  oblique  corner  of 
the  building,  and  draw  a pencil  line  on  the  under  side  of  the  block, 
along  the  upper  and  outer  edges  of  the  plates,  till  they  meet  at  the 
corner ; these  lines  will  be  the  bevel  of  the  backing  required.  And 
then  the  block  can  be  worked  off  to  the  lines,  and  a bevel  square  set 
to  the  angle  thus  formed,  by  which  the  hip  rafter  itself  can  then  be 
beveled. 


Length  of  the  Jack  Rafters. 

First . Of  those  on  the  long  side  of  the  building,  between  D and  L 
In  the  triangle  DFX,  the  jack  rafter/,  being  parallel  with  the  base  FI, 
has  the  same  proportion  to  FI  that  I )f  has  to  D1  (Geom.,  Props.  XX Y. 
& XXYIL) ; and  in  order  to  find  the  length  of  the  jack  rafters  coming 
between  I and  I),  we  have  only  to  divide  IF  by  the  number  of  rafters 
required  from  IF  to  the  corner  inclusively,  and  the  quotient  will  be 
the  difference  between  any  two  of  them,  and  will  equal  also  the  length 
of  the  shortest  one.  Since  it  is  14  feet  from  I to  the  corner  of  the 
building,  and  the  rafters  are  2 feet  apart,  it  will  require  7 rafters,  in- 
cluding IF.  We  therefore  divide  IF,  13  feet,  by  7,  and  have  1 foot 
lOf  inches  for  the  difference  between  IF  and  / and  also  between  /and 
e;  or,  what  is  equally  true,  we  can  take  this  number,  1 foot  lOf  inches, 
as  the  length  of  a;  double  this,  or  3 feet  inches,  for  the  length  of 
b;  three  times  this  number,  or  5 feet  67  inches,  for  the  length  of  c ; 7 
feet  5^  inches,  the  length  of  d;  9 feet  3|  inches,  the  length  of  e;  11 
feet  If  inches,  the  length  of  f;  and  13  feet  the  length  of  IF:  proving 
the  calculation  to  be  correct.  It  will  be  most  convenient  in  practice 
to  cut  the  longest  ones  first,  that  the  short  pieces  of  stuff  may  be 
worked  in  to  better  advantage. 

Second.  In  the  same  manner  obtain  the  difference  between  GF  and 


16 


CARPENTRY  MADE  EASY. 


&,  on  the  other  side  of  the  building,  bj  dividing  GF,  13  feet,  by  5,  tne 
number  of  rafters  required  in  GB ; 13  feet  divided  by  5 equals  2 feet 
7 J inches  ; this  is  at  once  the  length  of  g and  the  difference  between 
any  two  adjacent  rafters  between  B and  G. 

Third . Of  those  on  the  end  of  the  building.  Divide  the  length 
of  IIF  by  6,  which  is  the  number  of  rafters  required  between  HF 
and  each  corner  of  the  building,  including  HF.  HF  equals  the 
length  of  a common  rafter,  less  2 inches — the  bevel  of  the  hip  rafters 
— or  12  feet  10  inches,  which,  divided  by  6,  equals  2 feet  If  inches, 
the  length  of  each  of  the  short  rafters  on  the  end,  and  also  the  differ- 
ence between  HF  and  the  jack  rafters  on  each  side  of  HF. 

Side  Bevels  of  the  Jack  Rafters  on  the  Sides  of  the  Frame. 

It  is  evident  that,  if  the  roof  were  horizontal,  this  bevel  would  be 
found  on  the  square,  by  taking  half  the  width  of  the  building  on  the 
blade,  and  the  distance  from  the  corner  of  the  building  to  the  foot  of 
the  first  common  rafter  on  the  tongue ; but  when  the  roof  begins  to 
pitch,  this  bevel  will  be  too  short ; and  the  relation  of  the  pitch  of  the 
roof  to  the  length  of  the  common  rafter  is  such,  that  the  side  bevel  of 
the  jack  rafters  is  obtained  with  perfect  accuracy,  by  taking  the  length 
of  the  common  rafter  on  the  blade,  and  the  distance  from  D to  I and  from 
B to  0 , respectively , on  the  tongue — the  blade  will  give  the  side  bevels 
required. 

The  down  bevels  and  lower  end  bevels  of  these  jack  rafters  on  the 
sides  of  the  frame  are  the  same  as  those  of  the  common  rafters,  since 
they  have  the  same  pitch  that  the  common  rafters  have. 

Side  Bevels  of  the  Jack  Rafters  on  the  Slant  End  of  the  Frame. 

For  a similar  reason  assigned  above,  add  to  BG  the  gain  of  a common 
rafter  in  running  the  length  of  GB . Then  take  this  sum  on  the  blade  of  a 
square , and  half  the  width  of  the  building  on  the  tongue,  and  it  will  give 
the  side  bevel  of  the  end  rafters  which  are  nearest  BG.  In  like  man- 
ner, add  to  DI  the  gain  of  a common  rafter  in  running  the  length  of 
DI.  Take  this  sum  on  the  blade,  and  half  the  width  of  the  building 
on  the  tongue,  and  it  will  give  the  bevel  required  of  those  rafters 
nearest  DI. 

For  illustration,  BG  equals  10  feet ; as  the  common  rafters  are  13 
feet  long,  in  running  12  feet,  they  gain  an  inch  in  running  a foot.  So, 
in  running  10  feet,  a rafter  would  gain  10  inches.  We  therefore  take 
t,  certain  proportional  part  of  10  feet  10  inches  on  the  blade  of  a 


TRAPEZOIDAL  HIP  ROOFS. 


77 


square,  say  lOf  inches,  and  the  same  proportional  of  half  the  width  of 
the  building,  say  12  inches,  on  the  tongue ; the  bevel  of  the  blade  is 
the  side  bevel  for  all  those  jack  rafters  on  the  end  of  the  frame  which 
are  nearest  GB,  or  which  rest  against  BF. 

So,  also,  DI  equals  14  feet,  to  which  add  14  inches;  hence,  take 
15|  inches  on  the  blade,  and  12  inches  on  the  tongue ; the  bevel  on 
the  blade  will  be  the  side  bevels  of  those  jack  rafters  on  the  end  of 
the  frame  nearest  DI,  which  rest  against  the  hip  rafter  DF. 

Down  Bevel  of  the  Jack  Rafters  on  the  Beveled  End  of  the  Frame. 

In  order  to  obtain  these  bevels  and  the  corresponding  lower  end 
bevels  with  perfect  accuracy,  it  is  necessary  to  obtain  them  for  each 
rafter  separately,  for  there  are  no  two  of  them  which  have  the  same 
pitch.  In  order  to  ascertain  with  ease  what  this  pitch  and  the  corres- 
ponding bevels  are,  it  is  necessary  to  suppose  all  these  end  rafters  to 
be  produced  beyond  where  they  now  rest  on  the  hip  rafters,  and  all 
to  rest  upon  a common  ridge  pole  PR,  which  is  drawn  through  the 
point  F,  and  extends  on  a level  with  F directly  over  the  points  G and 
I.  These  rafters  would  then  have  each  one  the  same  pitch  they  now 
have,  since  we  have  not  supposed  their  direction  changed,  but  their 
length  only.  Now,  however,  the  rise  is  5 feet  for  each  one,  and  their 
run  can  easily  be  computed  from  the  length  of  BG ; for  the  run  of  the 
one  nearest  B is  4 inches  more  than  BG,  or  10  feet  4 inches ; that  of 
the  next  one,  4 inches  more,  or  10  feet  8 inches ; &c. 

Now,  in  order  to  obtain  the  down  bevel  and  the  lower  end  bevel  of 
any  rafter,  we  take  the  run  on  the  blade  of  a square,  and  the  rise  on 
the  tongue. 

The  measurements  of  these  bevels  are  therefore  as  follows : 

5 inches  on  the  tongue  for  each  of  them,  and  10 J inches,  10f  inches, 
II  inches,  11 J inches,  1 1 1 inches,  12  inches,  12 J inches,  12§  inches, 
13  inches,  13J  inches,  13§  inches,  respectively,  on  the  blade. 


PLATE  20. 


OCTAGONAL  AND  HEXAGONAL  ROOFS. 

Plate  20,  Fig.  1,  is  designed  to  exhibit  the  proper  mode  of  framing 
the  roof  of  a building,  the  ground  plan  of  which  is  a regular  octagon. 
This  style  of  building  having  become  quite  common,  a Rafter  1 able 
(No.  3)  has  been  prepared,  which  will  be  found  very  useful  and  con- 
venient, as  it  gives  at  one  view  the  precise  lengths  of  the  longest  hip 
rafters  and  the  longest  jack  rafters.  In  the  introduction  or  explana- 
tion of  the  table,  full  instruction  and  demonstrations  are  given  for 
enabling  the  intelligent  mechanic  to  test  these  calculations,  or  to  ex- 
tend and  apply  them  to  roofs  of  other  dimensions. 

Length  of  the  Hip  Balters, 

The  length  given  in  the  Table  is  that  of  the  first  pair,  and  is  cal- 
culated from  the  outer  and  upper  corner  of  the  plates,  at  their  inter- 
section with  each  other,  to  the  very  central  point  or  apex  of  the 
roof. 

One  of  the  first  pair  having  been  cut  off’ of  the  above  length  and  of 
the  proper  down  bevel,  the  length  of  the  second  pair  is  obtained  from 
this  one,  by  taking  off  from  this  one  half  its  thickness,  measured  back 
square  from  the  down  bevel ; and  the  length  of  the  third  and  fourth 
pairs,  by  taking  off  in  like  manner  two  thirds  its  thickness,  or  more 
accurately,  5]  of  its  thickness,  since  17  is  half  the  diagonal  of  a square, 
the  side  of  which  is  24.  (See  Prop.  XXI  V.,  Cor.) 

Bevels  of  the  Hip  Rafters. 

The  down  bevel , and  lower  end  bevel,  are  found  as  usual,  that  is,  by 
the  run  of  the  rafter  and  its  rise,  on  the  square.  (The  run  of  the  hip 
rafter  is  half  the  diagonal  width  of  the  building.  See  Explanation  of 
Octagonal  Hip  Roof  Table.) 

The  first  and  second  pairs  of  hip  rafters  have  no  side  bevel , since, 
in  raising  the  frame,  the  first  pair,  HI,  have  their  ends  resting  against 
each  other,  as  in  common  rafters.  The  second  pair,  AP,  are  next 
raised  at  right  angles  with  the  first,  and  their  ends  resting  square 
against  the  first.  The  third  and  fourth  pairs  have  their  side  bevels 
(78) 


fteo  T.etmharih  & Sou,  pil'd? 


OCTAGONAL  AND  HEXAGONAL  ROOFS. 


79 


cut  on  both  sides  of  the  upper  end ; and  that  bevel  is  found  by  taking 
the  length  of  the  rafter  and  its  run  on  the  square. 

The  backing  of  the  hip  rafter  is  obtained  on  the  square  by  taking  & 
of  its  rise  on  the  tongue,  and  its  length  on  :he  blade. 

Length  of  the  Jack  Rafters. 

The  length  oi  the  middle  jack  rafters  is  given  in  the  Table.  Having 
found  this,  in  order  to  obtain  the  length  of  the  shortest  one,  proceed  as 
directed  in  the  article  on  trapezoidal  hip  roofs;  that  is,  divide  the 
length  of  the  longest  one  by  the  number  of  rafters  required  between 
that  and  the  corner  inclusively— -the  quotient  will  be  the  length  of  the 
shortest  one,  and  also  the  difference  between  any  two  adjacent  ones. 

The  bevels  of  the  jack  rafters  are  obtained  on  the  same  prin- 
ciple as  those  in  common  hip  roofs. 

The  down  bevel  is  found  on  the  square  by  taking  the  length  and  the 
run  of  the  middle  jack  rafter;  and  the  side  bevel  by  taking  the  length 
©f  this  rafter  and  half  the  side  of  the  building. 

Fig.  2 represents  the  roof  of  a building,  the  ground  plan  of  which 
is  a regular  hexagon . 

Width  of  the  Building. 

In  every  regular  hexagon  the  side  is  equal  to  the  radius  of  the  cir- 
cumscribed circle ; and  the  diagonal  width  is  equal  to  the  diameter  of 
the  circumscribed  circle,  or  twice  the  side.  If  we  let  O'  represent  the 
foot  of  the  perpendicular  let  fall  from  O,  then  AB  and  BE  are  each 
equal  to  AO'  and  BO',  and  AD  is  equal  to  twice  AB.* 

Half  the  square  width  of  the  building  is  found  by  subtracting  from 
the  square  of  the  side,  the  square  of  half  the  side,  and  extracting  the 
square  root  of  the  difference:  since,  in  the  right-angled  triangle  CO'B, 
CO'2  =BO'2-— CB2.  The  side  of  this  building  being  supposed  to  be  20 
feet,  we  have  the  square  of  20=400,  and  the  square  of  '22o  = 100;  their 
difference  is  300  feet,  the  square  root  of  which  is  17.35  feet,  or  17  feet 
4,2  inches,  which  is  half  the  width  of  the  building,  or  the  run  of  the 
middle  jack  rafter 

The  Length  of  the  Rafter 

Must  be  computed  as  explained  in  the  Introduction  to  the  Balia 


• See  Geom.,  Prop.  XXXI. 


80 


CARPENTRY  MADE  EASY. 


Table ; that  is,  to  say,  every  rafter  is  the  hypotenuse  of  a right-anglevl 
triangle,  of  which  its  run  and  its  rise  are  the  other  two  sides.  The 
square  of  the  length  is  therefore  equal  to  the  sum  of  the  squares  of 
the  run  and  the  rise. 

In  this  Fig.  the  roof  is  supposed  to  rise  3 inches  to  the  foot.  The 
whole  rise  is  therefore  one  fourth  the  run  of  the  middle  jack  rafter,  or 
one  eighth  the  square  width  of  the  building,  or  4 feet  4.05  inches,  the 
square  of  which  is  2,709.2025  inches.  The  run  of  the  hip  rafter  we 
have  already  proved  to  be  equal  to  the  side  of  the  building,  or  20 
feet =240  inches,  the  square  of  which  is  57,600  inches.  The  square 
root  of  the  sum  of  the  above  two  numbers  is  245.56  inches,  or  20  feet 
5.56  inches,  which  is  the  length  of  the  hip  rafter. 

In  a similar  manner  the  length  of  the  middle  jack  rafters  is  found  to 
be  17  feet  10.26  inches.  The  length  of  the  shortest  jack  rafters  is 
obtained  in  the  same  manner  as  in  hip  roofs  generally. 

The  bevels  of  the  hip  jack  rafters  are  obtained  by  the  same  rule  as  in 
octagonal  roofs. 

The  backing  of  the  hip  rafters  is  found  on  the  square  by  taking  of 
its  rise  on  the  tongue,  and  its  length  on  the  blade. 


&?JT  o& 


1 


Plate  21. 


riieo.Leonharflt  Ic  SMn,Phil* 


PLATE  21. 


ROOFS  OF  BRICK  AND  STONE  BUILDINGS. 

Plate  21,  Fig.  1,  exhibits  a simple  and  excellent  mode  of  framing 
a roof  of  a moderate  span  of  from  25  to  85  feet,  designed  for  a stone 
or  brick  building.  The  iron  bolts,  by  which  the  feet  of  the  princi- 
pal rafters  are  secured  to  the  tie-beam,  are  } of  an  inch  in  di- 
ameter, and  the  supporting  rod  in  the  middle  is  1 inch  in  diameter. 
The  block  between  the  upper  ends  of  the  principal  rafters,  is  beveled 
to  suit  the  pitch  of  the  roof,  while  the  ends  of  the  principal  rafters 
are  square.  The  block  should  be  of  hard  wood,  and  placed  with  its 
grain  running  in  the  same  direction  as  that  of  the  rafters,  to  avoid 
shrinkage. 

By  means  of  the  supporting  rods,  any  proper  degree  of  camber  or 
crowning  can  be  given  to  the  tie-beam. 

The  bolts  and  rods  in  this  and  the  following  figures  are  left  rough 
in  the  Plates,  to  show  them  more  distinctly  ; but  if  the  rooms  beneath 
are  to  be  ceiled  or  plastered,  the  heads  of  the  bolts  should  be  counter- 
sunk into  the  beam,  and  the  nuts  screwed  upon  the  upper  ends. 

Fig.  2 represents  another  style-  of  roof,  with  two  braces  and  two 
additional  rods  to  each  bent,  and  also  a 2 inch  block  of  hard  wood 
about  2 feet  long,  placed  between  the  feet  of  the  braces.  The  grain 
of  this  block  should  also  run  parallel  with  the  beam. 

Length  and  Bevels  of  the  Braces. 

Suppose  the  distance  of  the  upper  end  of  the  brace  mortice  in  the 
principal  rafter  from  the  heel  of  that  rafter  to  be  five  ninths  of  the 
whole  length  of  that  rafter,  then  the  perpendicular  let  fall  from  the 
heel  of  the  brace  to  the  tie-beam  will  be  five  ninths  of  the  rise  of  the 
rafter — and  this  we  will  call  the  rise  of  the  brace  ; then  also  the  distance 
from  the  foot  of  this  perpendicular  to  the  toe  of  the  foot  of  the  brace, 
will  be  four  ninths  of  one  half  the  length  of  the  tie-beam  less  2 feet — 
one  foot  for  the  setting  back  of  the  principal  rafter,  and  one  foot  for 
half  the  block  at  the  foot  of  the  brace  : this  is  the  run  of  the  brace;  then 
the  sum  of  the  squares  of  the  rise  and, the  run  will  give  the  square  of 
6 (81; 


82 


CARPENTRY  MADE  EASY. 


the  length  of  the  brace,  the  square  root  of  which  will  be  that  length 
from  the  lower  toe  to  the  upper  heel. 

The  Span  of  this  Roof 

May  be  from  50  to  65  feet.  The  middle  rod  should  be  inches 
the  others  f of  an  inch. 

Dimensions  of  Timbers  for  Figs,  1 & 2. 

For  loth  frames — Rafters,  2 by  6 ; plates  and  purlin  plates,  6 by  8. 

For  each  bant  in  Fig.  1— Tie-beams,  8 by  10 ; principal  rafters,  7 by  9. 

For  each  bent  in  Fig.  2— Tie-beams,  10  by  12 ; principal  rafters,  7 by 
10 ; braces,  6 by  7. 

The  bents  in  each  frame  may  be  from  10  to  14  feet  apart. 

Fig.  3 represents  a simple  mode  of  framing  a roof  for  a shop  or 
foundry  where  it  is  required  to  have  a ventilator. 


Scale  S feet  to  tlm  inch. 


PLATE  22. 


Fig.  1 represents  a bent  of  a very  strong  roof,  designed  for  machine 
shops  and  other  buildings,  where  it  is  necessary  for  great  weights  or 
heavy  machinery  to  be  supported  by  them.  The  Author  of  this  work 
has  superintended  the  construction  of  roofs  of  this  kind  upon  the  ma- 
chine shops  for  the  Railroad  Works  at  Peoria,  111.,  and  at  Davenport, 
Iowa,  where  they  have  proved  sufficiently  strong  to  sustain  locomo- 
tive engines  weighing  more  than  twenty  tons,  when  hoisted  from  the 
ground,  and  suspended  by  chains  from  the  roof  for  repairs. 

The  width  of  the  building  represented  in  Fig.  1 is  50  feet ; the  prin- 
cipal rafter  is  set  back  a foot  from  the  end  of  the  tie-beam,  to  give 
room  for  the  wall  plate ; the  rise  of  the  roof  is  5 inches  to  the  foot. 
In  framing  roofs  of  this  kind,  the  supporting  rods  should  be  furnished 
before  commencing  the  frame : for  then  the  length  of  the  short  prin- 
cipal rafters  and  that  of  the  straining  beam,  can  be  regulated  or  pro 
portioned  according  to  the  length  of  the  rods.  It  is  best,  however, 
for  the  middle  rod  to  be  twice  the  length  of  the  short  ones,  reckoning 
from  the  upper  surface  of  the  beam  to  the  upper  surface  of  the  prin- 
cipal rafters,  and  allowing  1 foot  more  to  each  rod  for  the  thickness 
of  the  beam,  and  the  nut  and  washer.  For  example,  the  middle  rod 
is  11  feet  long,  and  the  short  ones  6 feet  each;  which,  after  allowing 
1 foot,  as  above  mentioned,  makes  the  length  of  the  long  one,  above 
the  work  side  of  the  beam,  twice  that  of  the  short  ones. 

The  length  of  the  rod  above  the  beam  is  the  rise  of  the  rafter,  and 
the  distance  from  the  centre  of  the  rod  to  the  foot  of  the  rafter  is  the 
run  of  the  rafter ; the  length  of  the  rafter  can  therefore  be  found  from 
the  Common  Rafter  Table. 

length  of  the  Straining  Beam. 

Add  the  run  of  the  short  principal  rafter  to  the  lower  end  bevel  of 
the  long  one ; subtract  this  sum  from  the  run  of  the  long  principal, 
and  the  difference  will  be  half  the  length  of  the  straining  beam. 

The  bolsters  under  the  ends  of  the  tie  beams  are  of  the  same  thick® 
ness  as  that,  and  about  5 feet  long. 

Fig.  2 is  in  every  respect  similar  to  Fig.  1,  with  the  exception 
of  the  long  principal  rafters  and  the  middle  supporting  rod  being 
omitted.  This  roof  is  suitable  for  blacksmith  shops  and  foundries, 
as  it  is  also  capable  of  sustaining  great  weights,  and  may  be  very  con- 
venient for  the  safe  storing  of  unwrought  iron  bars  upon  the  beams. 

(83) 


PLATE  23. 


This  plate  exhibits  sections  of  three  roofs,  of  different  dimensions, 
but  similar  to  each  other  in  style.  This  style  is  ancient,  and,  no  doubt, 
has  been  proved  of  sufficient  strength ; but  it  is  not  recommended  for 
convenience  or  economy,  except  where  labor  is  cheap,  timber  plenti- 
ful, and  iron  scarce. 

In  Fig.  1,  the  post  in  the  middle  is  called  the  king  post,  and  the 
other  two  queen  posts . The  tie  beam  is  secured  to  their  feet  by  iron 
straps ; and  braces  extend  in  pairs  from  the  posts  to  the  principal 
rafters,  as  represented  in  the  Plate.  The  heads  of  the  posts  are  bev- 
eled to  correspond  with  the  pitch  of  the  roof,  and  the  ends  of  the  prin- 
cipal rafters  are  left  square. 

The  manner  of  scarfing  the  tie  beam  is  represented  immediately 
below. 

Fig.  2 is  very  similar  in  design  to  Fig.  1,  in  the  preceding  Plate, 
the  principal  difference  consisting  in  having  king  and  queen  posts 
instead  of  supporting  rods. 

Fig.  3 shows  the  ridge  pole  supported  by  braces  from  the  ends  of 
the  straining  beam.  The  king  post  is,  therefore,  omitted;  and  the 
space  between  the  queen  posts  may  be  appropriated  for  an  attic  cham- 
ber. The  queen  posts  are  let  into  the  tie-beam  an  inch  or  more,  to 
prevent  displacement  by  the  lateral  pressure  of  the  braces. 

(84) 


7* 


Plato  24 


PLATE  24. 


Plate  24  exhibits  several  designs  for  roofs  in  a new  and  improved 
style,  particularly  adapted  to  those  of  a great  span,  as  they  may  be 
safely  extended  to  a very  considerable  width,  with  less  increase  of 
weight,  and  less  proportionate  expense,  than  any  of  the  older  styles. 
The  principle  on  which  they  are  constructed  is  essentially  the  same 
as  that  of  the  Howe  Bridge.  The  braces  are  square  at  the  ends,  the 
hard  wood  blocks  between  them  being  beveled  and  placed  as  described 
in  the  foregoing  Plates.  Each  truss  of  this  frame  supports  a purlin 
post  and  plate,  as  represented. 

These  roofs  are  easily  made  nearly  flat,  and  thereby  adapted  to 
metallic  covering,  by  carrying  the  walls  above  the  tie  beams  to  any 
desired  height,  without  altering  the  pitch  of  the  principal  rafters, 
which  ought  to  have  a rise  of  at  least  4 inches  to  the  foot,  to  give  a 
sufficient  brace  to  the  upper  chord  or  straining  beam. 

Fig.  1 is  represented  with  counter-braces ; and 

Fig.  2 without  them.  The  counter-braces  do  not  add  any  thing  to 
the  mere  support  of  the  roof,  and  are  entirely  unnecessary  in  frames 
of  churches,  or  other  public  buildings,  where  there  is  no  jar ; but  they 
may  very  properly  be  used  in  mill  frames,  or  other  buildings  designed 
for  heavy  machinery. 


m 


PLATE  25. 


This  Plate  exhibits  two  plans  for  roofs  of  the  same  style  as  the  last, 
but  of  simpler  construction,  and  designed  for  a shorter  span. 

In  Fig.  2 the  middle  truss  is  omitted,  to  afford  room  for  an  attio 
chamber. 

(86) 


Plate  25. 


Tb,p o .Tconhar (l  t *<  Sun.  _PUila 


» 


Plate  26 


H)  ft 


PLATE  26. 


This  Plate  exhibits  several  designs  of  Gothic  roofs,  the  manner  of 
framing  which  is  sufficiently  indicated  by  the  Plate. 

Fig.  1 is  constructed  entirely  of  wood. 

^ ! 

Fig.  2 of  wood,  strengthened  with  iron  straps  and  bolts ; md  i 

Fig.  3 with  still  less  wood,  but  supported  by  iron  rods ; and,  un- 
doubtedly, the  strongest  roof  of  the  three. 

The  first  is,  however,  a neat,  cheap,  and  very  simple  plan,  and  suf- 
ficiently strong  for  a roof  having  a steep  pitch,  and  ©f  not  more  than 
40  feet  span. 


(87) 


PLATE  27. 


Plate  27  represents  two  designs  for  church  roofs,  with  arched  or 
vaulted  naves. 

In  Fig.  1 the  arch  is  formed  of  2 inch  planks,  from  6 to  8 inches 
wide,  after  being  wrought  into  the  proper  curve.  These  planks  are 
doubled,  so  as  to  break  joints,  and  firmly  spiked  together.  Lighter 
arches,  of  similar  construction,  are  sprung,  at  a distance  of  16  inches 
apart,  between  the  bents,  for  supporting  the  lathing. 

In  Fig.  2 the  arch  is  formed  of  3 inch  planks,  10  to  12  inches  wide, 
and  made  in  three  sections,  and  spiked  to  the  braces,  as  represented. 

Note. — The  foregoing  designs  for  roofs  have  been  selected  from  more  than  a hundred 
drafts  in  the  Author’s  possession,  and  are  believed  to  be  the  best  selection  ever  offered 
to  the  public  eye.  The  number  could  have  been  increased  with  ease  to  an  indefinite  ex- 
tent; but  it  has  been  deemed  necessary  to  insert  those  only  which  are  at  once  excellent 
and  practicable,  and  which  combine  the  latest  improvements. 

(88) 


Plate  27. 


I 


Theo.  Leoiihardt  & So  a.  Phil* 


■ 


PLATE  28. 


Plate  28  exhibits  the  frame- work  of  a church  spire,  85  feet  high 
aliove  the  tie  beam,  or  cross  timber  of  the  roof.-  This  is  framed  square 
as  far  as  the  top  of  the  second  section,  above  which  it  is  octagonal.  It 
will  be  found  most  convenient  to  frame  and  raise  the  square  portion 
first;  then  to  frame  the  octagonal  portion,  or  spire  proper,  before  rais- 
ing it : in  the  first  place  letting  the  feet  of  the  8 hip  rafters  of  the  spire, 
each  of  which  is  48  feet  long,  rest  upon  the  tie  beam  and  joists  of  the 
main  building.  The  top  of  the  spire  can,  in  that  situation,  be  conve- 
niently finished  and  painted,  after  which  it  may  be  raised  half  way  to 
its  place,  when  the  lower  portion  can  be  finished  as  far  down  as  the 
top  of  the  third  section.  The  spire  should  then  be  ? aised  and  bolted 
to  its  place,  by  bolts  at  the  top  of  the  second  section  at  AB,  and  also 
at  the  feet  of  the  hip  rafters  at  CD.  The  third  section  can  then  be 
built  around  the  base  of  the  spire  proper ; or  the  spire  can  be  finished, 
as  such,  to  the  top  of  the  second  section,  dispensing  with  the  third, 
just  as  the  taste  or  ability  of  the  parties  shall  determine. 

Fig.  2 presents  a horizontal  view  of  the  top  of  the  first  section. 

Fig.  8 is  a horizontal  view  of  the  top  of  the  second  section,  after  the 
spire  is  bolted  to  its  place. 

The  lateral  braces  in  the  spire  are  halved  together,  at  their  intersec- 
tion with  each  other,  and  beveled  and  spiked  to  the  hip  rafters  at  the 
ends.  These  braces  may  be  dispensed  with  on  a low  spire. 

A conical  finish  can  be  given  to  the  spire  above  the  sections,  by 
making  the  outside  edges  of  the  cross  timbers  circular. 

The  bevels  of  the  hip  rafters  are  obtained  in  the  usual  manner  for 
octagonal  roofs,  as  described  in  Plate  20 

Note. — In  most  cases  the  side  of  an  octagon  is  given  as  the  basis  of  calculation  in  find- 
ing the  width  and  other  dimensions;  but  in  spires  like  this,  where  the  lower  portion  is 
square,  we  are  required  to  find  the  side  from  a given  width.  The  second  section  in  this 
steeple,  within  which  the  octagonal  spire  is  to  be  bolted,  is  supposed  to  be  12  feet  square 
outside ; and  the  posts  being  8 inches  square,  the  width  of  the  octagon  at  the  top  of  this 
section,  as  represented  in  Fig.  3,  is  10  feet  8 inches,  and  its  side  is  4 feet  5.02  inches, 
as  demonstrated  in  the  explanation  of  the  Table  for  Octagonal  Roofs  (No.  3). 

The  side  of  any  other  octagon  may  be  found  from  this  by  proportion,  since  all  regular 
octagons  are  similar  figures,  and  their  sides  are  to  each  other  as  their  widths,  and,  con- 
versely their  widths  are  to  each  other  as  their  sides.-— See  Explanation  of  Table  No.  3. 

(89) 


PLATE  29. 


Plate  29  exhibits  the  plan  of  a large  dome  of  60  oi  75  feet  span, 
built  upon  a strong  circular  stone  or  brick  wall.  In  constructing  this 
roof,  there  are  four  bents  framed,  like  the  one  exhibited  in  Fig.  1,  all 
intersecting  each  other  beneath  the  king  post  at  the  centre.  The  tie 
beams  in  the  first  and  second  bents  are  of  full  length,  and  halved 
together ; those  in  the  third  and  fourth  bents  are  in  half  lengths,  and 
mitred  to  the  intersection  of  the  first  and  second. 

The  King  Post 

Has  eight  faces,  and  on  each  face  two  braces ; one  large  brace  from 
the  top  of  the  post  to  the  end  of  the  tie  beam,  and  one  small  brace 
from  the  bottom  of  the  post  to  the  middle  of  the  large  brace.  These 
four  tie  beams  are  supported  by  eight  posts,  extending  from  the  top 
of  the  main  wall  to  the  ends  of  the  beams,  and  3ach  one  braced  as 
represented  in  the  figure. 

Two  circular  arches,  constructed  of  planks,  as  described  in  Plate 
27,  are  then  sprung,  one  above  and  one  below  each  bent,  as  repre- 
sented in  the  figure.  Between  each  of  these  four  arches,  three  others 
are  constructed,  supported  by  short  timbers,  framed  into  the  ties  of 
the  tie  beam,  as  represented  in  Fig.  2. 

Fig.  2 is  a horizontal  section  of  the  dome,  drawn  through  it  at  the 
main  tie  beam  AB,  which  corresponds  with  AB,  in  Fig.  1. 

Fig.  3 is  a horizontal  view  of  the  apex  of  the  dome,  where  all  the 
32  arches  intersect  each  other,  showing  the  mode  of  beveling  them  at 
their  intersection. 

(90) 


Plate  29. 


Tire  a Leonhard  t Ik  Son,  PliU% 


PART  III. 


Plate  30. 


BRIDGE  BUILDING. 


PLATE  30. 

STRAINING  BEAM  BRIDGES. 

Plate  80,  Fig.  1 represents  a straining  learn  bridge  of  30  feet  span,  de 
signed  for  i.  common  highway.  The  stringers  or  main  timbers  are  35 
feet  long,  extending  over  each  abutment  to  a distance  of  2 J feet.  The 
straining  beam  is  equal  in  length  to  J of  the  span,  or  10  feet.  The 
supporting  rods  are  8 feet  2 inches  long : 1 foot  is  allowed  for  the 
thickness  of  the  stringer,  10  inches  for  the  needle  beam,  and  4 inches 
nut  head  and  washers;  leaving  6 feet  as  the  rise  of  the  brace,  or  the 
distance  of  the  top  of  the  straining  beam  from  the  top  of  the  stringer. 

The  length  of  the  brace  can  therefore  be  found,  as  usual,  by  extracting 
the  square  root  of  the  sum  of  the  squares  of  the  run  and  the  rise. 

Bevels. 

The  bevel  at  the  foot  of  the  brace  is  like  that  at  the  foot  of  a rafter, 
and  is  obtained  in  the  same  manner.  The  bevel  at  the  upper  end  of 
the  brace  and  the  bevel  of  the  straining  beam  are  equal  to  each  other, 
and  are  each  equal  to  half  that  of  a rafter  of  the  same  rise  and  run. 

Fig.  3 exhibits  the  horizontal  plan  of  the  floor  timbers,  and  the 
manner  of  laying  both  the  joists  and  the  planks. 

A moderate  degree  of  camber  should  be  given  to  every  bridge  of 
this  kind,  by  screwing  up  the  supporting  rods. 

Bill  of  Timber, 

2 Stringers,  12  by  12,  in.  35  feet  long.  Board  measure  =840  feet. 


4 Braces,  8 by  10,  “ 

12 

u 

tt 

it 

a 

=160 

ti 

2 Straining  beams  8 by  10,  “ 

10 

it 

it 

it 

a 

=133 

a 

2 Wall  plates,  10  by  12,  “ 

16 

a 

it 

it 

u 

=320 

a 

2 Needle  beams,  8 by  10,  “ 

18 

it 

' i 

a 

a 

=240 

a 

6 Joists,  3 by  10,  “ 

12 

a 

it 

ti 

a 

=150 

.i 

(93) 

94 


CARPENTRY  MADE  EASY. 


6 Joists,  3 by  10,  22  feet  long,  Board  measure =245  feet 

932  feet,  2-incb  planks  “ M =932  “ 


Total  timber,  B.  M.  3020  “ 

Bill  of  Iron. 

4 Supporting  rods,  1J  in.  diameter,  8 ft.  2 in.  long,  each  34  J lbs. =138  lbs. 

8 Washers,  4 lbs.  each;  and  4 nuts  1 lb.  each,  = 36  “ 

4 Bolts,  1 in.  diameter  22  in.  long,  each  5J  lbs.  =22  11 

8 Washers,  1 lb.  each,  and  4 nuts  £ lb.  each,  =11  “ 

40  lbs.  spikes,  =40  “* 


247  " 

Estimate  of  Cost 

3020  feet  lumber,  @ $15  per  M.  =$45.30 

247  lbs.  iron,  @ 7c.  “ lb.  =17.29 

Workmanship,  @ $10  “ M.  Board  measure.  = 30.20 


Total  cost,  $92.79 


Fig.  2.  In  respect  to  this  bridge,  it  is  only  necessary  to  say  that  it  is 
constructed  upon  the  same  principle  as  the  former ; the  difference  being 
caused  only  by  the  increase  of  the  span,  and  this  difference  being  suffi- 
ciently represented  by  the  Plate. 

In  raising  the  former  of  these  bridges,  no  false  work  or  temporary 
supports  are  needed,  but  for  this  one  they  may  be. 

Concerning  the  economy  and  durability  of  these  bridges,  it  may  be 
proper  to  observe  that  they  are  comparatively  simple  and  cheap ; and 
they  are  also  sufficiently  strong,  so  long  as  the  supports  maintain  their 
vertical  position.  But  this  plan  has  two  objections. 

1.  The  absence  of  side  braces  induces  a leaning  or  twisting  of  the 
braces,  caused  by  their  pressure  toward  each  other ; and  when  this 
twisting , or  torsion  as  it  is  often  called,  has  once  commenced,  it  cannot 
well  be  remedied.  It  may,  however,  be  guarded  against,  to  a certain 
extent,  by  such  a modification  of  the  design  as  will  allow  of  two  sup- 
porting rods  at  each  end  of  the  needle  beams — these  rods  being  crossed ; 
one  passing  inside  of  the  stringer,  and  the  other  at  some  distance  outside 
of  it,  toward  the  end  of  the  needle  beam. 

2.  The  absence  of  counter  braces  exposes  the  bridge  to  injury  from 
vibration;  which  is  specially  destructive  to  the  stone- work  of  the 
abutments,  the  repeated  jars  being  almost  sure  to  break  the  mortar 
and  loosen  the  stones.  The  use  of  a wall-plate  serves  in  some  degree 
to  obviate  this  objection ; and  in  case  of  the  bridge  being  supported  by 
trestles,  it  disappears. 


t 


Floor.  Fhzn 


Plate:}! 


D 


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n 


[HI 


tL 


0 


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Ehfco.  Lsouliaidt  & Sm..1‘1u]u 


PLATE  31. 


Tins  ondge  is  more  expensive  and  more  durable  than  th(>se  before 
represented,  as  it  is  also  less  liable  to  the  objections  mentioned  con- 
cerning  them.  The  counter  braces  of  this  bridge  are  sufficient  to  pre- 
vent injurious  effects  from  vibrations ; and  the  size  of  the  posts,  or  up- 
right ties,  when  secured  by  straps  of  iron,  as  represented,  will  also 
prevent  the  torsion  or  twisting  of  the  braces,  to  which  the  others  are 
liable.  The  manner  of  framing  this  bridge  is  sufficiently  indicated 
by  the  Plate ; and  the  lengths  and  bevels  of  the  braces  are  obtained  as 
usual. 


Bill  of  Timber, 


2 String  pieces, 

12  by  12  in. 

65  feet  long 

2 Straining  beams, 

12  by  12  “ 

18 

a 

tt 

4 Long  braces, 

12  by  12  “ 

26 

u 

it 

4 Short  end  braces, 

10  by  12  “ 

15 

a 

a 

4 Middle  braces, 

10  by  12  “ 

10 

a 

tt 

4 Counter  braces, 

10  by  12  “ 

9 

a 

a 

4 Long  posts, 

12  by  12  “ 

10 

u 

u 

2 Middle  posts, 

12  by  12  “ 

9 

u 

4 Short  posts, 

12  by  12  “ 

6 

a 

a 

2 Wall  plates, 

6 by  12  “ 

18 

u 

ti 

5 Needle  beams, 

10  by  10  « 

20 

ti 

it 

12  Joists, 

3 by  10  “ 

24 

ti 

a 

6 Joists, 

8 by  10  “ 

18 

a 

tt 

2000  feet,  B.  M.,  of  floor  plank, 

16 

a 

a 

1560 

feet 

432 

u 

1248 

a 

600 

n 

400 

it 

360 

a 

480 

it 

216 

it 

288 

tt 

216 

u 

833 

a 

720 

ti 

270 

tt 

2000 

it 

(95) 


9623 


PLATE  32. 


This  Plate  presents  a view  Af  a bridge  offered  as  an  improvement 
of  the  Howe  Bridge,  of  a moderate  span,  by  shortening  the  upper 
chord,  and  bracing  the  ends  of  it  in  the  same  manner  as  in  a straining 
beam  bridge.  In  the  Howe  Bridge,  the  upper  chord  is  of  the  same 
length  as  the  lower  one,  and  the  braces  and  counter  braces  are  placed 
in  a uniform  manner  throughout  the  entire  length.  In  the  plan  repre 
sented  in  this  Plate,  by  reducing  the  length  of  the  upper  chord  to  the 
limit  of  a single  piece  of  timber,  it  is  proposed  to  secure,  at  least,  an 
equal  degree  of  strength  to  the  ordinary  Howe  Bridge,  and  at  the 
same  time  to  effect  economy  in  both  material  and  labor. 

. The  ends  of  the  braces  are  left  square,  and  the  proper  bevels  are 
made  upon  the  angle  blocks,  which  are  of  hard  wood  or  of  cast  iron, 
and  are  let  into  the  chords  to  the  depth  of  1 inch  or  more. 

The  main  braces  all  lean  inward  toward  the  centre  of  the  span,  and 
are  double,  passing  one  outside  and  one  inside  of  the  counter  braces, 
which  are  single,  leaning  in  the  opposite  direction  from  the  centre  to- 
ward the  ends,  each  brace  passing  between  each  pair  of  main  braces, 
and  are  all  three  bolted  together  at  their  intersection. 

The  lower  chords  in  each  truss,  or  each  side,  are  three  in  number, 
and  bolted  together  in  the  most  firm  manner  possible.  Hard  wood 
keys,  2 inches  thick,  6 inches  wide,  and  12  inches  long,  are  inserted 
on  each  side  of  every  joint,  and  at  certain  intervals  even  where  there 
are  no  joints.  These  keys  are  let  into  the  chords  only  about  three 
fourths  of  an  inch  on  each  side,  leaving  a half  inch  space  between  the 
chords  for  the  free  circulation  of  air. 


Bill  of  Timber. 


2 Upper  chord  pieces, 

10  by  14  in.  54  feet  long=1260  feet. 

4 Long  end  braces, 

10  by  14  “ 22 

u 

“ =1027 

a 

4 Short  “ 11 

6- by  6 “ 12 

a 

“ = 144 

a 

4 Short  end  counter  braces, 

4 by  6 “ 12 

a 

“ = 96 

a 

32  Middle  main  braces, 

5 by  7 “ 13 

a 

“ =1212 

a 

12  Counter  braces, 

4 by  7 “ 13 

a 

“ = 364 

a 

6 Lower  chord  pieces, 

6 by  12  “ 31 

a 

" =1116 

i 

^ a it  a 

6 by  12  1 10 

u 

“ = 960 

u 

RBIDGE  BUILDING. 


97 


4:  Lower  chord  pieces, 


6 by  12  in.  30  feet  long—  720  feet. 


4.  a u a 

6 by  12  “ 23 

u 

“ = 552  “ 

2 Wall  plates, 

8 by  12  “ 20 

a 

" = 320  “ 

32  Joists, 

4 by  12  “ 18 

u 

“ =2304  “ 

10  Lateral  braces, 

4 by  6 “ 24 

u 

“ = 480  “ 

3000  feet,  B.  M.,  floor  plank, 

3000  “ 

Bill  of  Iron 

4 Middle  support,  rods, 

1 in. 

diam.,  12  ft.  2 in. 

long,  32  lbs.—  128  Ibsi 

8 Next  to  middle  u 

“ 12  “ 2 “ 

a 

51  “ = 408  “ 

16  End  rods, 

u 

“ 12  “ 2 “ 

a 

73  “ =1168  “ 

8 Short  end  rods, 

1* 

“ 6 “ 1 “ 

a 

36J  “ = 292  “ 

4 Long  cross  rods, 

1 

00 

cc 

ti 

48$  “ = 194  “ 

4 End  bolts, 

1 

a 2 " 0 “ 

a 

5£“  = 21  “ 

24  Lower  chord  bolts, 

1 

“ 22  “ 

a 

5 “ = 120  " 

16  Brace  bolts, 

1 

a 16  “ 

a 

2 “ = 32  “ 

72  Nuts  for  supporting  rods, 

2 “ = 144  •' 

18  Plates  “ “ 

“ | in.  thick,  4 w.,  14  long,  12  11  — 216  u 

18  “ “ " 

u 

a 4 w9, 19  long,  16  11  = 288  “ 

96  Washers, 

1 “ = 96  “ 

48  Nuts  for  small  bolts, 

1 “ = 48  “ 

Note.-t The  cost  of  labor  in  constructing  this  bridge  I®  estimated  at  $11,00  per  thou- 
sand,  B.  M.,  of  the  timber  required. 


7 


PLATE  33. 


TRESTLE  BRIDGES 

This  Plate  exhibits  the  design  of  a bridge  supported  from  below  ; 
and,  for  a moderate  span,  it  is  one  in  which  the  important  elements 
of  simplicity,  strength,  and  durability,  are  well  combined. 

The  plan  of  this  bridge  is  so  simple,  as  to  require  little  further  ex- 
planation than  the  inspection  of  the  Plate.  It  will  be  perceived  that 
the  bearings  are  JO  feet  apart,  and  that  the  braces  are  framed  to  cor- 
respond. The  cn  t$s  timbers  are  extended  out  several  feet  on  each 
side,  to  give  room  for  bracing  the  hand-rail. 

This  bridge  is  supported  by  trestles;  and  the  Plate  represents  the 
manner  of  framing  the  end  ones  and  the  middle  one.  It  is  of  the 
utmost  importance  that  the  embankments  behind  the  end  trestles  are 
perfectly  solid,  as  on  their  firmness  depends  the  whole  strength  of  the 
bridge. 


Bill  of  Timber  for  One  Span. 


4 String  pieces, 

12  by  12  in.,  28  feet  long, 

=1344  feet 

2 Straining  beams, 

12  by  16  “ 30 

u 

a 

= 960 

it 

4 Long  braces, 

10  by  10  “ 24 

a 

it 

= 800 

it 

4 Short  braces, 

10  by  10  “ 14 

tt 

a 

- 467 

U 

6 Cross  timbers, 

6 by  12  “ 24 

a 

t 

= 864 

It 

12  Hand-rail  posts, 

6 by  6 “ 4i 

a 

tt 

= 162 

it 

104  feet  (lineal)  hand  railing, 

6 by  6 “ 

= 312 

it 

12  Hand-rail  braces, 

3 by  4 “ 4i 

tt 

it 

= 54 

it 

12  Joists, 

3 by  10  “ 20 

a 

a 

= 600 

It 

6 

3 by  10  “ 11 

it 

tt 

= 165 

tt 

1660  feet,  B.  M.,  floor  plank, 

=1660 

tl 

Total 

7388 

u 

Bill  of  Timber  for  the  Two  End  Trestles. 

4 Posts, 

12  by  12  in.,  13  feet  long, 

= 624  feel 

8 Studs, 

6 by  12  “ 13 

it 

a 

= 624 

tt 

4 “ 

6 by  12  “ 8 

u 

t 

= 192 

4 

(98) 


BRIDGE  BUILDING. 


99 


6 by  12  in.,  6 feet  long,  = 144  feet 

12  by  12  “ 52  “ “ =1248  “ 

12  by  12  “ 20  41  “ = 480  “ 

wood  plank  for  supporting  embankm’t,=1000  n 

4812  w 


Bill  of  Timber  for  Middle  Trestle. 


8 Posts, 

12  by  12 

in., 

18  feet  long,  B.M., 

= 468  feet 

1 Mud  sill, 

12  by  12 

a 

80  “ 

u 

44 

= 360 

a 

1 Cap, 

12  by  12 

u 

20  “ 

4 i 

<4 

= 240 

u 

4 Post  head  braces, 

4 by  4 

it 

5}  “ 

it 

it 

= 30 

u 

2 “ foot  braces, 

8 by  8 

a 

8 “ 

44 

a 

= 90 

h 

1188 

u 

For  the  two  end  trestles, 

4312 

it 

Total  of  the  three  trestles,  Board  measure, 

5500 

it 

4 Studs, 

2 Mud  sills, 

2 Caps, 

1000  ft.,  2 in.  bard 


PLATES  34  & 35. 


Plates  34  and  35  represent  a strong  trestle  bridge,  sucb  as  is  often 
used  for  rail-roads  in  crossing  small  streams  and  ravines,  where  the 
banks  are  high,  and  where  there  is  little  danger  from  ice.  The  Author 
of  this  work  has  constructed  bridges  of  this  kind  at  Spring  Creek,  Bu- 
reau Co.,  and  at  Nettle  Creek,  Grundy  Co.,  on  the  Chicago  and  Bock 
Island  Rail-road ; and  one  on  the  plank  road,  between  Peru  and  La 
Salle,  in  La  Salle  Co.,  111.-— the  last  with  posts,  51  feet  high. 

In  framing  the  trestles,  the  posts  are  framed  into  the  sills  and  caps  as 
usual ; but  the  braces  are  bolted  upon  the  outside  with  inch  bolts.  The 
outside  lower  braces  of  the  trestles,  marked  in  the  plan  C,  C,  have  1 ' 
foot  run  to  2 feet  rise ; the  posts  of  the  trestles  at  A are  set  in  such  a 
manner  as  to  act  as  braces,  having  1 foot  run  to  4 feet  rise.  The  hori • 
zontal  lateral  braces  are  also  laid  and  bolted  between  the  longitudinal 
timbers  and  cross  timbers,  without  being  framed  into  them.  The 
lower  longitudinal  timbers  are  let  into  the  posts  to  the  depth  of  2 
inches,  and  lapped  across  the  posts,  one  on  one  side,  and  the  other  on 
the  other  side,  where  they  are  bolted  to  the  posts  and  to  each  other. 

The  bearings  are  ten  feet  apart , and  each  bearing  is  supported  either 
by  a post  or  a brace ; these  braces  are  framed  to  a 10  feet  rise  and  a 9 
feet  run,  and  the  upper  ends  are  bolted  to  the  longitudinal  timbers,  as 
represented  in  the  Plate. 

A bill  of  timber  and  iron , which  is  here  subjoined,  will  assist  the  me- 
chanic in  framing  a bridge  of  this  kind  more  than  any  extended  de- 
scription could  do.  (The  small  letters  in  the  Plate  refer  to  the  bolts.) 


Bill  of  Timber. 


6 Sills, 

1 Sill, 

1 “ 

18  Posts, 
3 “ 


12  by  12  in.,  33  feet  long,  B 
12  by  12  “ 18  “ “ 

12  by  12  “ 16  “ “ 


M.,  —2376  feet 
= 216  “ 


- 192  “ 


3 


8 1 <jr  cross  timbers, 
100) 


12  by  12  “ 25  “ 
12  by  12  “ 12  “ 

12  by  12  “ 10  " 

12  by  12  “ 19  “ 

12  by  12  “ 11  “ 

6 by  12  “ 19  “ 


=5400  “ 
= 432  “ 
= 360  “ 
=1368  “ 
= 264  “ 
= 342  “ 


Tbeu . J,eo»)\arSt  S Scm.Thi'.? 


c . k 


Floor  Ft  an. 


Cross  Section 


BRIDGE  BUILDING. 


101 


6 Lower  longitudinal  timbers,  6 by  12 

in., 

35  feet  long, 

18  Braces, 

8 by  10 

u 

14 

it 

it 

6 “ 

8 by  10 

a 

15 

it 

tt 

6 String  pieces, 

9 by  12 

it 

27 

tt 

u 

8 “ “ 

9 by  12 

u 

32 

It 

tt 

9 Cross  timbers, 

10  by  12 

a 

19 

tt 

u 

16  Lateral  braces, 

4 by  8 

u 

27 

it 

a 

8 Rail  stringers, 

12  by  14 

u 

30 

it 

it 

4 Bolsters, 

12  by  12 

u 

6 

it 

u 

6 Cross  braces, 

4 by  8 

u 

14 

tt 

tt 

6 “ “ 

4 by  8 

it 

19 

tt 

tt 

2 “ “ 

4 by  8 

u 

16 

it 

tt 

2 “ “ 

4 by  8 

a 

18 

It 

tt 

Total,  Board  measure, 

Bill  of  Iron. 

6 Bolts  (letter  a), 

32  in.  long, 

1 in.  diam., 

12  “ ( “ 5), 

36  “ 

n 

a 

12  « ( “ c), 

18  “ 

u 

a 

18  “ ( M d), 

22  “ 

u 

tt 

3 “ ( “ e), 

22  tl 

u 

tt 

27  “ ( “ /), 

31  “ 

it 

tt 

18  “ ( “ g), 

26  “ 

tt 

tt 

8 “ ( “ h), 

28  " 

it 

tt 

294  Heads,  nuts,  and  washers,  @ 1 lb.  each, 

18  Bolts  (letter  i), 

20  in.  long, 

1 

in. 

diam., 

6 “ ( « j), 

11  “ 

u 

n 

20  “ ( “ k), 

18  u 

it 

u 

2 « ( “ 0, 

22  M 

it 

u 

138  Bolt-heads,  nuts,  and  washers,  @ f lb.  each, 


Total  of  iron, 


=1260  feet 
=1680  “ 
= 600  “ 
=1458  “ 
= 864  “ 
=1710  “ 
=1152  “ 
=8860  “ 
= 288  “ 
= 224  “ 
= 304  “ 
= 85  “ 
= 96  “ 

24,031  “ 


= 421  lbs 
= 95|  “ 
= 24  “ 

= 87*  “ 
= 15  “ 

=185  “ 

=103*  “ 
= 50  “ 

=294  “ 

= 45  “ 

= 8|  “ 
= 45  “ 

- 5i  ‘ 

=103  ‘ 

1100*  « 


PLATE  30. 


ARCHED  TRUSS  BRIDGES 

This  Plate  represents  a design  of  a Burr  Bridge  without  counter 
braces,  but  combined  with  an  arch  beam.  This  mode  of  construction 
is  designed  either  for  railroad  bridges,  or  for  common  road  bridges  of 
a great  span.  If  wanted  for  a common  road,  and  the  span  be  not 
more  than  150  feet,  the  arch  beam  may  be  safely  dispensed  with ; and 
in  that  case,  counter  braces  should  be  introduced  * but  if  the  bridge 
be  designed  for  a railroad,  the  arch  beam  should  never  be  omitted. 

The  panels  of  bridges  of  this  kind  ought  never  to  be  as  great  in 
extension  as  in  height  between  chords;  or,  in  other  words,  the  rise  of 
the  braces  should  always  be  greater  than  their  run ; and  practically, 
it  is  expensive  and  inconvenient  to  extend  the  panels  more  than  12 
or  14  feet.  In  all  bridges  of  this  kind,  the  greatest  strain  upon  the 
braces  is  at  the  end  of  the  span ; and  it  will  be  most  proper  to  use  the 
best  and  largest  pieces  of  timber  for  the  end  braces,  and  those  of  in- 
ferior quality,  if  such  must  be  used  somewhere,  should  be  placed  in 
the  middle. 

The  posts  should  be  sized  down  at  the  lower  end,  where  they  pass 
through  the  lower  chord,  to  about  6 inches  in  thickness ; the  chord 
pieces  should  also  be  cut  out  to  the  depth  of  1 inch  on  each  side  of 
the  post,  and  both  locked  into  the  post  in  the  firmest  possible  manner, 
in  order  to  resist  the  thrust  of  the  brace.  The  post  should  also  be 
boxed  into  the  upper  chord  not  less  than  1 inch. 

In  scarfing  the  lower  chord  pieces,  they  must  be  so  arranged  that 
only  one  splice  be  made  at  the  same  place ; and  if  the  bolts  which  pass 
through  the  scarfing  extend  also  .through  both  lower  chord  pieces, 
(the  short  piece  inserted  to  lock  the  joint  being  of  just  sufficient  thick- 
ness to  fill  the  space  between  the  two  chord  pieces),  it  would  be  still 
better  than  that  plan  represented  in  the  Plate. 

It  will  be  found  necessary,  in  a bridge  of  this  kind,  to  make  the 
main  braces  at  least  1 J inches  longer  than  the  exact  calculation  would 
require,  in  order  to  produce  the  necessary  camber,  and  to  guard 
against  the  settling  of  the  centre  of  the  span  below  the  general  level, 
which  will  be  likely  tc  happen  if  not  guarded  against,  from  the  com- 
(102) 


BtttD&E  BUtLbltfQi 


10B 


pression  and  shrinkage  of  the  timber,  and  which  would  tnatefially 
weaken  the  bridge ; and  whatever  camber  the  bridge  is  designed  to 
have,  must  be  given  to  it  on  its  first  erection,  before  the  false  works 
are  removed,  since  the  camber  cannot  afterward  be  increased  as  it  can 
be  in  most  of  the  bridges  represented  on  the  preceding  Plates,  where 
supporting  rods,  in  those  plans,  occupy  the  place  of  the  posts  in  this. 

For  tloor  plan,  see  Plate  37. 


Bill  of  Timber  for  One  Span. 


2 Wall  plates, 

10  by  12 

in.. 

, 22  feet  long,  B.  M.=  440  feet. 

8 Lower  chord  pieces, 

7 by  14 

u 

34 

a 

tt 

“ = 2286  “ 

g u tt  tt 

7 by  14 

a 

43 

it 

tt 

“ = 2809  “ 

4 Upper  “ “ 

11  by  11 

tt 

36 

tt 

it 

“ = 1452  “ 

^ tt  tt  it 

11  by  11 

a 

44 

a 

tt 

“ = 1774  “ 

46  Floor  beams, 

4 by  12 

tt 

20 

a 

a 

« = 3680  ^ 

60  Arch  pieces, 

7 by  10 

a 

33 

a 

n 

" =11550  u 

12  Queen  posts, 

11  by  15 

tt 

19 

a 

it 

" = 3135 

8 “ “ 

11  by  14 

tt 

19 

it 

u 

“ = 1950  “ 

8 “ “ 

11  by  13 

it 

19 

u 

a 

« = 1811  u 

2 Centre  posts, 

11  by  16 

tt 

19 

tt 

tt 

u = 557  “ 

4 Arch  braces, 

12  by  14 

tt 

18 

it 

tt 

M = 1008  “ 

28  Panel  girths, 

6 by  8 

a 

11 

it 

tt 

« = 1232  “ 

12  Scarfing  blocks, 

6 by  14 

it 

5 

a 

tt 

“ = 420  “ 

16  Tie  beams, 

10  by  10 

u 

19 

a 

tt 

M = 2533  " 

32  Knee  braces, 

4 by  6 

it 

6 

a 

tt 

“ = 884  “ 

18  Lateral  braces, 

4 by  8 

tt 

24 

it 

tt 

= 1152  “ 

7000  feet  3 inch  floor  plank, 

14 

a 

u 

^ = 7000  “ 

Bill  of  Iron. 


9 Gross  rods,  1 in.  diam.  19  feet  3 inches  long  =460  .ba. 


22  Arch  bolts,  1 a 

24  Splice  bolts,  1 a 

30  Post  bolts,  1 “ 

60  Arch  bolts,  f “ 

Head,  nut,  and  washer  to  each  bolt 


2 

a 9 

a 

it 

=160 

tt 

1 

foot  8 

a 

tt 

= 106 

:t 

1 

“ 8 

a 

it 

=132 

tt 

2 

feet  0 

tt 

tt 

= 179 

tt 

lb. 

each, 

345 

a 

PLATE  37. 


This  Plate  represents  a first-class  Howe  Bridge  for  a Bailroad,  and 
as  one  of  similar  principles  and  construction  has  been  fully  explained  in 
the  description  of  Plate  32,  page  96,  it  will  only  be  necessary  here  to 
add  the  bills  of  timber  and  iron,  in  detail. 

For  a common  highway,  a bridge  of  this  length  will  admit  of  some 
modifications,  as  it  is  heavier  than  necessary. 

The  cost  of  a bridge  of  this  kind,  in  the  State  of  Illinois,  would  be 
about  $30  per  lineal  foot. 


Bill  of  Timber  for  One  Span  Howe  Bridge,  150  feet  long. 


4 Wall  plates, 

10  by  12 

in., 

20  feet  long, 

= 800  feet. 

8 Bolsters, 

10  by  14 

« 

10 

(( 

1C 

= 933 

tt 

12  Lower  chords, 

6 by  14 

tt 

42 

u 

a 

=3528 

tt 

2 “ “ 

6 by  14 

tt 

32 

a 

a 

448 

tt 

2 “ “ 

6 by  14 

tt 

22 

a 

a 

= 308 

ft 

2 “ “ 

6 by  14 

ft 

12 

it 

tt 

= 168 

ft 

12  “ “ 

7 by  14 

a 

42 

a 

it 

=4116 

ft 

2 “ “ 

7 by  14 

tt 

32 

a 

tt 

= 522 

tt 

2 “ “ 

7 by  14 

tt 

22 

u 

a 

= 359 

it 

2 « “ 

7 by  14 

tt 

12 

a 

a 

= 196 

tt 

12  Upper  " 

6 by  12 

tt 

42 

a 

a 

=3024 

ft 

2 “ “ 

6 by  12 

tt 

32 

a 

a 

= 384 

tt 

2 “ “ 

6 by  12 

it 

22 

a 

a 

= 264 

tt 

2 “ “ 

6 by  12 

tt 

12 

a 

tt 

= 144 

tt 

12  “ “ 

7 by  12 

a 

42 

a 

tt 

=3328 

tt 

2 “ “ 

7 by  12 

« 

32 

a 

ft 

= 448 

tt 

2 “ “ 

7 by  12 

« 

22 

a 

tt 

= 308 

tt 

2 “ “ 

7 by  12 

(C 

12 

a 

tt 

= 168 

tt 

8 Main  braces, 

10  by  11 

u 

24 

a 

tt 

=1760 

ft 

8 “ “ 

10  by  10 

<( 

24 

a 

tt 

=1600 

tt 

8 “ “ 

10  by  9 

a 

24 

a 

ft 

=1440 

tt 

8 “ “ 

10  by  8 

a 

24 

a 

it 

=1280 

tt 

8 “ “ 

10  by  7 

(( 

24 

a 

ft 

=1120 

tt 

16  “ “ 

10  by  6 

a 

24 

a 

tt 

=1920 

a 

28  Counter  “ 

6 by  8 

a 

24 

a 

tt 

=2688 

tt 

56  Lateral  “ 

6 by  6 

u 

18 

a 

tt 

=3024 

a 

46  Cross  floor  timbers, 

7 by  14 

(( 

20 

a 

tt 

=7513 

tt 

12  Bail  stringers, 

10  by  12 

(6 

26 

it 

tt 

=3120 

tt 

4 Struts, 

6 by  12 

(( 

16 

tt 

tt 

= 320 

tt 

16  End  supports, 
(104) 

6 by  12 

a 

20 

it 

tt 

=1600 

tt 

J 


Plate  ;‘>7. 


Th«*o.icm»lv«rfli  & Soa.Vhi]* 


BRIDGE  BUILDING. 


105 


700  ft.  Oak  for  clamps,  3 by  14  in.,  12  feet  long,  = 700  feet. 

500  “ u Packing  blocks,  3 by  14  a 12  " “ = 500  “ 

500  “ “ Upper  « “ 3 by  12  “ 12  « “ = 500  « 


48,531 

Bill  of  Wrought  Iron. 

8 Supporting  rods,  If  in.  diam.. 

22  feet 

8 

in. 

long, 

=1961 

lbs. 

40  “ “ 

ij  “ 

22 

tt 

8 

tt 

tt 

=5436 

tt 

24  “ “ 

it  “ 

22 

tt 

8 

a 

tt 

=2720 

tt 

24  “ “ 

u « 

22 

« 

8 

a 

tt 

=2310 

tt 

6 " “ 

1 “ 

22 

« 

8 

a 

tt 

= 380 

a 

8 “ “ 

1 “ 

23 

a 

6 

a 

tt 

= 517 

tt 

30  Lateral  “ 

1 “ 

19 

« 

6 

u 

tt 

=1580 

tt 

180  Chord  bolts, 

1 “ . 

2 

n 

8 

tt 

tt 

= 720 

tt 

24  Gibs,  5 holes, 

1 by  6 in., 

3 

u 

0 

tt 

ft 

=1440 

tt 

24  “ 3 “ 

1 by  5}  “ 

2 

<( 

6 

tt 

tt 

=1080 

tt 

4 “ 3 “ 

f by  5 “ 

2 

u 

6 

tt 

ft 

= 127 

tt 

666  lbs.  Nuts, 

= 566 

tt 

18,837 

Bill  of  Castings. 

4 Half  angle  blocks. 

94 

lbs, 

. each, 

= 376  lbs. 

24  Angle  blocks, 

178 

tt 

=4272 

tt 

24  “ « 

148 

tt 

=3552 

tt 

4 tt  tt 

132 

a 

= 528 

tt 

60  Lateral  angle  blocks. 

21 

tt 

=1260 

tt 

136  Washers, 

4 in.  diam. 

3 

tt 

= 408 

tt 

360  Chord  bolt  washers,  3 “ 

tt 

= 450 

it 

10,846 

1st  Set  of  Rods,  next  to  abutments,  2 rods,  If  in.  iron,  3 rods,  1 J in.  iron. 


2d  “ 

tt 

u 

tt 

2 

tt 

ij 

tt 

3 “ If  “ 

3d  “ 

tt 

tt 

tt 

2 

tt 

i* 

tt 

3 “ IJ  « 

4th  “ 

tt 

tt 

tt 

3 

a 

H 

ft 

5th  “ 

tt 

tt 

tt 

3 

tt 

if 

it 

6th  “ 

tt 

it 

ft 

3 

tt 

H 

tt 

Center  “ 

tt 

tt 

ft 

3 

tt 

l 

tt 

Note.— It  will  be  seen  that  the  first  3 panels,  nest  to  the  abutments,  are  supported  by  5 rods 
each,  and  that  the  outside  and  inside  rods  are  beyond  the  chords,  passing  through  tubes  in  the 
angle-blocks,  also  the  heaviest  main  braces  are  used  next  to  the  abutments,  growing  lighter  as 
they  approach  the  ©enter. 


PLATE  38. 


This  bridge  is  similar,  in  its  general  principles  of  construction,  to 
the  one  represented  in  Plate  36 ; but  is  quite  different  in  its  minor  de- 
tails, being  much  heavier  and  stronger,  as  well  as  more  expensive. 
The  main  differences  are  these : Counter  braces  are  employed  in  this 
biidge,  which  are  omitted  in  the  other ; this  has  two  sets  of  posts  and 
main  braces,  and  but  one  arch  beam  to  each  truss,  while  the  other 
bridge  has  two  arch  beams  and  one  set  of  posts  and  braces;  the  chord 
pieces  in  this  bridge,  instead  of  being  placed  side  by  side,  with  their 
edges  vertical,  with  an  open  space  between  them  for  the  circulation 
of  air,  are  placed  one  upon  the  other,  with  their  edges  horizontal,  and 
their  surfaces  in  close  contact. 

The  upper  chord  is  in  three  sections,  and  the  lower  chord  and  the 
arch  beam  are  each  in  four  sections ; each  chord  piece  and  arch  piece 
being  6 inches  deep  and  12  inches  wide ; making  the  combined  upper 
chord  12  by  18  inches,  and  the  combined  lower  chord  and  arch  beam 
each  12  by  24  inches. 

The  foot  of  the  arch  beam  rests  upon  a cast  iron  shoe,  secured  by 
iron  straps  to  each  of  the  lower  chord  pieces  ; each  shoe  having  four 
flanges,  and  each  flange  beveled  to  fit  the  square  end  of  each  section 
of  the  arch  beam. 

There  is  one  set  of  counter  braces  to  each  truss,  each  counter  brace 
passing  between  each  pair  of  main  braces,  to  which  it  is  bolted  at  their 
intersection.  The  foot  of  each  counter  brace  rests  upon  an  angle  block 
fixed  upon  the  lower  chord,  at  the  foot  of  each  pair  of  posts,  and  the 
upper  end  of  each  counter  brace  rests  against  the  arch  beam  at  its  inter- 
section with  the  next  pair  of  posts.  A key  is  inserted,  however,  be- 
tween the  upper  end  of  each  counter  brace  and  the  arch  beam,  by  means 
of  which  the  whole  structure  can  be  kept  tight,  and  the  relative  strain 
upon  the  arch  beam  and  the  chords  can,  to  some  extent,  be  regulated 
and  proportioned.  Each  pair  of  posts  is  bolted  together  with  four 
bolts — one  above,  and  one  below  each  chord. 

Bridges  of  this  style  are  in  extensive  use  on  the  New  York  and 
Erie  Bail-road,  where  they  have  been  proved  to  be  of  great  strength 
and  stability. 

(106) 


GENERAL  PRINCIPLES  OF  BRIDGE  BUILDING. 


In  concluding  this  Part  of  the  work,  it  is  proper  to  bring  together 
into  one  place  the  most  important  principles  and  most  useful  hints  to 
practical  builders,  which  we  have  been  able  to  gather,  either  from  the 
study  of  other  works,*  or  from  the  lessons  of  our  own  experience. 

Size  of  Timber  and  Iron  required  to  enable  a Bridge  of  a given  Span  to 

sustain  a given  Load. 

The  most  proper  way  of  ascertaining  the  resisting  powers  of  timber 
and  iron  is  by  actual  experiment;  and  it  has  been  found  by  such  ex 
periment,  that  the  greatest  safe  strain  for  sound  timber  is  about  1,000 
lbs.  per  square  inch,  measured  on  the  square  end  of  the  timber,  the 
strain  being  one  of  either  extension  or  compression,  but  applied  in  the 
direction  of  the  grain  of  the  wood.  It  has  also  been  ascertained  by 
experiment,  that  the  greatest  safe  tensile  strain , as  it  is  technically 
called,  that  is  the  lifting  or  supporting  strain,  of  large  wrought  iron 
rods,  is  10,000  lbs.  per  square  inch.  Small  wire  or  nail  rods,  manu 
factured  with  more  care,  and  of  the  best  materials,  can,  undoubtedly, 
sustain  a much  greater  weight  than  this. 

In  proportioning  the  different  parts  of  a bridge,  however,  it  is  cus> 
tomary  and  expedient  to  allow  a considerable  excess  of  strength  in 
favor  of  stability.  The  deterioration^  of  timber,  caused  by  age,  must 
be  taken  into  the  account ; for  after  a wooden  bridge  has  been  in  use 
for  some  years,  it  becomes  much  weaker  than  when  first  erected. 

The  weight  of  the  bridge  itself  must  also  be  considered  in  deter- 
mining the  load  which  it  is  able  to  sustain,  and  this  weight  it  is  con? 
sidered  safe  to  assume  at  35  lbs.  to  the  cubic  foot  of  timber  employed. 
If  the  quantity  of  timber  in  a given  bridge  is  equal  to  30  cubic  feet 
to  every  foot  in  length,  as  is  asserted  by  Haupt  to  be  the  case  with 
the  average  of  the  Howe  Bridges  on  the  Pennsylvania  Kail-road,  then 
the  weight  of  the  structure  would  be  1050  lbs.  per  lineal  foot,  or  a 


* Many  of  these  remarks  are  condensed  and  simplified  from  the  work  on  “ Bridge 
Construction,”  by  Herman  Haupt,  A.  M. — D.  Appleton  & Co.,  New  York,  a work  more 
especially  designed  for  the  use  of  engineers  than  for  practical  builders,  yet  one  wh  ch 
we  commend  to  all  persons  interested  in  this  part  of  Carpentry. 


(107) 


108 


CARPENTRY  MADE  EASY. 


little  more  than  half  a ton  per  foot  for  the  weight  of  the  timber,  ex- 
clusive of  the  iron. 

The  greatest  load  that  can  be  brought  upon  a rail-road  bridge, 
with  a single  track,  is  when  several  locomotive  engines  of  the  first 
class,  weighing  about  one  ton  per  foot  in  length,  are  attached  to- 
gether. So  that  the  greatest  strain  upon  such  a bridge,  including  both 
its  own  weight  and  the  weight  of  the  load,  is  a little  more  than  a ton 
and  a half  per  foot.  What,  then,  must  be  the  dimensions  of  the  tim- 
ber to  resist  this  strain  ? 

The  Strain  upon  the  Chords. 

When  a beam  is  supported  at  the  ends  and  loaded  in  tne  middle 
until  it  breaks,  it  is  observed  that  the  fibres  in  the  lower  portion  of 
the  fracture  are  broken  by  being  extended  or  pulled  violently  apart, 
and  that  those  on  the  upper  portion  are  broken  by  being  compressed 
or  jammed  violently  together.  In  theory,  this  compression  is  said  to 
be  equal  to  the  expansion ; that  is,  that  it  will  require  an  equal  force 
to  tear  the  fibres  apart  as  to  break  them  by  forcing  them  together, 
and  the  neutral  axis  in  the  beam,  or  the  line  where  there  is  neither 
sufficient  expansion  nor  compression  to  break  the  fibres  of  the  timber, 
is  said  to  be  in  the  middle  of  the  beam.  But  it  is  doubtful  whether 
facts  will  warrant  this  conclusion.  Common  observation  would  lead 
most  persons  to  the  opinion  that  timber  has  a greater  power  to  resist 
compression  than  it  has  to  resist  expansion,  and  to  this  opinion  we 
are  ourselves  inclined ; but  for  the  present  purposes  it  will  be  suffi- 
ciently accurate  to  be  governed  by  the  theory  usually  adopted  by  en- 
gineers, as  stated. 

The  power  of  a bridge  to  sustain  a load,  and  to  resist  the  various 
strains  upon  it,  may  be  compared  to  that  of  the  beam  supported  at  the 
ends — the  strain  on  the  upper  chord  being  one  of  compression,  and 
that  on  the  lower  chord  one  of  extension ; and  the  strain  on  both  being 
greatest  in  the  middle  of  the  span,  and  diminishing  toward  the  ends. 
When  the  beam  is  laid  over  several  supports,  its  strength  for  a given 
interval  is  much  greater  than  when  simply  supported  at  the  ends. 
The  same  principle  is  applicable  to  bridges ; and  when  several  spans 
occur  in  succession,  it  is  of  great  advantage  to  continue  the  upper  and 
lower  chords  across  the  piers. 

The  greatest  strain  on  the  upper  chord  being  in  the  middle  of  the 
span,  is  equal  to  that  force  which,  being  applied  horizontally,  would 
sustain  one  half  the  span  with  its  load  were  the  other  half  to  be  re 


BRIDGE  BUILDING. 


109 


moved.  In  order  to  ascertain  this  force,  multiply  half  the  span  and 
its  load  by  one  fourth  its  length,  and  divide  that  product  by  its  height, 
measured  from  centre  to  centre  of  the  upper  and  lower  chords. 

For  example,  if  the  length  of  a span  be  160  feet,  and  the  height  of 
the  truss  be  16  feet  from  centre  to  centre  of  upper  and  lower  chords, 

nd  the  weight  of  the  loaded  bridge  be  1 J tons  to  the  lineal  foot,  the 
greatest  strain  upon  the  upper  chord  would  be  expressed  by  the 
product  of  120  tons  multiplied  by  40,  and  the  product  divided  by  16 ; 
which  gives  300  tons,  or  600,000  lbs.  as  the  result.  The  reason  of 
multiplying  the  weight  of  half  the  loaded  span  by  40  is,  because  40 
feet  is  the  middle  of  the  half-span,  or  its  centre  of  gravity ; and  the 
reason  for  dividing  its  product  by  16  is,  because  that  is  the  width  of 
the  truss ; and  the  wider  the  truss,  the  greater  leverage  there  is,  and 
the  less  strain,  for  the  same  reason  that  a thick  beam  is  stronger  than 

t ' ° 

a flat  one,  as  there  is  less  strain  on  the  upper  and  lower  surfaces  of 
the  thick  beam  from  the  same  weight  than  in  a flat  one.  Then,  as 
each  square  inch  is  able  to  resist  1000  lbs.,  there  must  be  600  square 
inches  in  the  end  section  of  the  upper  chords,  in  order  to  enable  them 
to  sustain  the  weight  required,  or  300  inches  in  the  upper  chord  of 
each  truss.  If,  therefore,  each  chord  is  12  inches  deep,  it  must  be  25 
inches  wide ; hence,  three  chord  pieces,  12  by  8 J inches,  will  contain 
the  requisite  material 

The  strain  on  the  lower  chord  is  at  least  equal  to  that  on  the  upper 
one ; but  the  timbers  being  in  several  pieces,  and  the  strain  being  one 
of  extension,  the  joints  are  opened,  and  the  whole  strength  of  the  tim- 
ber is  not  available ; while  in  the  upper  chord  the  strain  is  one  of  com- 
pression, and  the  joints  being  pressed  together,  causes  no  loss  to  the 
resisting  force  of  the  timber.  There  must,  therefore,  be  an  additional 
quantity  of  timber  in  the  lower  chord ; and  each  piece  should  be  suffi- 
ciently long  to  extend  through  four  panels,  so  that  there  can  be  three 
whole  timbers  and  a joint  in  each  panel. 

From  the  same  data,  similar  calculations  can  easily  be  made  fci 
estimating  the  strain  and  fixing  the  dimensions  of  the  other  timbers. 


PART  IV 


PLATE  39. 


This  Plate  is  introduced  here  to  demonstrate  more  fully  the  proper 
way  of  obtaining  the  backing  on  hip  rafters:  continued  from  page  71, 
where  we  say  that  “ this  is  found  on  the  square  by  taking  the  length  of 
the  hip  rafter  on  the  blade,  and  the  rise  of  the  roof  on  the  tongue ; the 
bevel  on  the  tongue  is  the  backing  required.”  To  find  the  proportional 
length  of  the  hip  rafter,  we  take  a two  foot  rule  and  measure  the 
diagonal  distance  on  a square,  by  placing  one  end  of  the  rule  upon  the 
17  inch  mark  of  the  blade,  and  the  other  end  upon  that  mark  of  the 
tongue  which  corresponds  with  the  rise  of  the  roof  to  the  foot ; this  gives 
the  proportional  length  of  the  hip  rafter ; then  apply  this  length  to  the 
blade  of  a square  and  the  rise  of  the  roof  to  the  tongue,  as  above  stated 
on  page  71,  and  the  true  bevel  will  be  obtained. 

To  illustrate  and  make  it  more  plain,  we  will  find  the  backing  of  a 
hip  rafter  on  a roof  which  rises  5 inches  to  the  foot.  On  applying  the 
two  foot  rule  to  the  square  at  the  17  inch  mark  on  the  blade,  and  at  the 
5 inch  mark  on  the  tongue,  we  find  the  diagonal  distance  to  be  17f 
inches ; taking  this  last  measure  on  the  blade  and  the  5 inch  mark  on 
the  tongue  and  applying  them  to  a straight  edge,  the  tongue  will  indi- 
cate the  backing  required,  as  shown  at  A,  Fig,  1. 

The  intersection  of  the  two  squares,  each  set  in  the  same  manner  as 
above  described,  shows  the  backing  on  both  sides  of  the  rafter  at  B,  and 
a bevel  square  set  to  the  proper  angle  is  exhibited  at  C. 

If  the  rise  of  the  roof  is  4 inches  to  the  foot,  we  proceed  as  before, 
and  find  the  diagonal  or  proportional  length  of  the  hip  rafter  to  be  17J 
inches,  which  we  apply  to  the  straight  edge  on  the  blade  and  the  4 inch 
mark  on  the  tongue,  and  find  the  proper  backing  for  a 4 inch  pitch.  In 
a similar  manner  a 10  inch  pitch  gives  a diagonal  of  19f  inches,  and  in 
fine,  this  rule  is  correct  for  any  pitch. 

The  reason  why  we  use  the  17  inch  mark  on  the  blade  of  the  square 
for  this  purpose  is  because  the  diagonal  of  a square  foot  is  16^0  inches, 
to  which  17  inches  is  as  near  as  we  can  practically  work.  See  pages  132 
and  133,  where  these  principles  are  fully  explained. 

Fig.  2 exhibits  a practical  method  of  working  off  the  corners  of  square 
timber  to  make  it  octagonal  or  eight  square.  In  this  Fig.  TM  represents 
any  piece  of  timber  not  more  than  2 feet  square,  on  which  we  place  the 

(111) 


112 


CARPENTRY  MADE  EASY. 


square  as  represented,  with  the  heel  at  one  edge  and  the  end  of  the  blade 
at  the  other ; then  prick  the  timber  at  the  7 and  17  inch  marks  accur- 
ately, and  set  the  gauge  accordingly  and  gauge  each  face,  then  take  off 
the  corners  to  these  gauge  marks.  This  method  is  sufficiently  accurate 
for  common  purposes,  the  error  being  only  T$u  of  an  inch,  for  the  square 
of  7 is  49,  and  twice  49  is  98,  and  this  is  the  face  of  the  corner  which  is 
taken  off,  the  middle  face  which  is  left  being  10  inches,  of  which  the 
square  is  100.  Should  the  timber  be  larger  than  24  inches  square,  take 

of  the  width  of  each  face,  and  proceed  as  before. 

Fig.  3 exhibits  another  method.  Let  AB  represent  the  diagonal 
drawn  across  the  end  of  a square  stick  of  timber,  and  let  the  dotted  lines 
intersecting  at  the  center  be  arcs  drawn  from  each  corner  with  a radius 
of  half  the  diagonal,  then  take  off  the  corners  to  the  points  where  the 
arcs  cut  the  faces,  and  the  octagon  is  exact. 

Mechanics,  as  a rule,  square  their  work  by  the  numbers  6,  8,  and  10, 
which  is  correct,  yet  only  a few  of  them  know  the  reason  why  it  is  so. 
It  is  simply  because  these  numbers  represent  the  three  sides  of  a right- 
angled  triangle  of  which  10  is  the  hypotenuse,  the  square  of  which  is 
equal  to  the  sum  of  the  squares  of  the  other  two  sides ; i.  e.,  6 times  6 is 
36,  and  8 times  8 is  64,  and  the  sum  of  36  and  64  is  100,  which  is  also 
the  square  of  10. 


; - , ■ 


PLATE  40. 


This  Plate  is  introduced  here  to  show  the  proper  method  of  bracing 
between  posts  that  stand  battering  or  inclined,  as  in  a derrick  for  a wind- 
mill, In  Fig.  1 the  posts  incline  or  batter  two  inches  to  the  foot  in  rise ; 
the  bevel  of  the  posts  at  the  top  and  bottom  is  obtained  as  usual,  by- 
taking  12  inches  on  the  blade  of  the  square,  and  2 inches  on  the  tongue, 
and  also  for  the  ends  of  the  girder,  which  has  the  same  bevel. 

The  braces  on  the  lower  side  of  the  girder  must  be  set  as  many  inches 
nearer  the  shoulder  of  the  girder  as  the  post  gains  in  battering.  If  the 
post  batters  two  inches  in  rising  one  foot,  and  if  the  brace  is  framed  on 
a regular  run  of  three  feet,  the  toe  of  the  brace  will  be  six  inches  nearer 
the  shoulder  than  it  would  be  were  the  post  to  stand  plumb.  Hence  the 
true  distance  of  the  toe  of  the  lower  brace  from  the  shoulder  of  the 
girder  is  2 feet  6 inches,  and  on  the  upper  side  the  brace  will  be  3 feet 
6 inches,  for  a similar  reason. 

To  find  the  distance  on  the  post  for  the  toe  of  the  brace,  turn  to  the 
common  rafter  table,  page  130,  and  take  the  length  of  a rafter  for  a 
building  which  is  double  the  width  of  the  run  of  the  brace,  and  that  will 
be  the  true  distance  from  the  girder  to  the  toe  of  the  brace  either  above 
or  below  the  girder.  For  example,  in  Fig.  1 the  braces  on  the  girder 
are  each  framed  for  a regular  run  of  3 feet  and  the  posts  battering  2 
inches,  we  turn  to  the  rafter  table  and  find  that,  for  a building  6 feet 
wide  and  with  a rise  of  2 inches  to  the  foot,  the  length  of  the  rafter  is 
3 feet  of  an  inch,  which  is  the  exact  distance  of  the  toe  of  the  brace 
from  the  girder. 

Practically  this  distance  may  be  found  by  the  square,  thus : Apply  the 
square  to  the  post,  placing  the  end  of  the  blade  at  the  shoulder  of  the 
girder  and  the  4 inch  mark  on  the  tongue  either  above  or  below ; then 
at  this  latter  point  apply  the  12  inch  mark  of  the  blade  again  and  the 
2 inch  mark  on  the  tongue  still  higher  or  lower  upon  the  post,  and  this 
last  point  will  be  the  true  place  for  the  toe  of  the  brace,  and  on  meas~ 
uring  the  distance,  it  will  be  found  to  be  3 feet  TV©  of  an  inch,  as  before. 

Should  the  battering  or  inclination  of  the  posts  be  either  more  or  less 
than  2 inches  to  the  foot,  then  the  same  principles  will  apply,  by  vary- 
ing the  use  of  the  tables  and  the  square  accordingly. 

8 


(113) 


114 


CARPENTRY  MARE  EASY. 


Bevels  of  the  Braces. 

There  are  three  different  bevels  of  the  braces:  1.  Those  upon  the 
girder  are  all  regular  and  at  an  angle  of  45°.  2.  Those  above  the  girder 
are  as  much  more  #ban  45°  as  the  girder  is  out  of  square ; and  3.  Those 
below  the  girder  are  as  much  less  than  45°  as  those  above  are  more,  the 
on?  being  the  complement  of  the  other.  The  most  convenient  way  to 
find  these  bevels  is  first  to  cut  the  regular  bevels  at  the  lower  end  of  the 
braces,  as  at  D in  Fig.  2,  then  measure  its  length  from  D to  E,  then 
mark  or  cut  the  upper  end  at  the  same  bevel  parallel  to  the  first,  and 
then  apply  to  this  mark  or  section  a bevel  square  which  has  been  set  to 
the  exact  bevel  of  the  end  of  the  girder ; then  one  end  of  the  blade  will 
show  the  bevel  of  the  lower  brace,  as  at  E,  and  by  turning  the  bevel 
square  upside  down,  the  other  end  of  the  blade  will  show  the  bevel  of 
the  upper  brace,  as  at  B,  Fig.  3. 

Height  of  the  Girder. 

To  ascertain  the  slant  height  of  the  girder  from  the  sill  upon  the 
posts,  suppose  we  want  the  perpendicular  height  to  be  10  feet,  then  from 
the  common  rafter  table  as  before,  we  find  the  length  of  a rafter  for  a 
building  20  feet  wide,  with  a rise  of  2 inches  to  the  foot.  This  is  found 
to  be  10  feet  lv/^o  inches,  which  is  the  slant  height  of  the  girder  from 
the  sill.  Or  take  5 lengths  of  the  blade  of  the  square,  applied  as  before, 
with  the  4 inch  mark  of  the  tongue  upon  the  post,  and  the  lust  point 
will  be  the  true  distance  of  the  girder  upon  the  post  from  the  sill. 

Length  of  the  Girder  and  Cap. 

To  find  the  length  of  the  girder,  multiply  its  perpendicular  height  in 
feet,  by  the  battering  of  both  the  posts  to  a foot,  in  inches ; the  product 
will  be  in  inches  to  be  subtracted  from  the  length  of  the  sill  between  the 
posts.  In  this  example  we  subtract  40  inches  from  the  inner  length  of 
the  sill,  because  10  feet,  the  height  of  the  girder,  multiplied  by  4 inches, 
the  battering  of  both  the  posts  to  a foot,  produces  40  inches. 

In  the  same  manner  find  the  length  of  the  cap  by  subtracting  the 
product  of  its  height,  multiplied  by  the  battering  of  the  two  posts,  from 
the  outer  length  ,of  the  sill. 

The  Long  Braces. 

To  frame  the  long  braces  between  the  posts,  as  shown  in  Fig.  1,  first 
find  the  distance  between  the  posts  at  the  bottom  of  the  braces  in  the 
same  manner  as  before  in  finding  the  length  of  the  girder,  then  as  the 
braces  should  be  framed  on  a regular  run,  or  at  an  angle  of  45c,  we  see 


BATTERING  POSTS. 


lift 


that  if  the  posts  were  7 feet  apart  at  the  dotted  line  AA',  the  brace 
would  rise  just  6 feet  and  run  6 feet  at  its  intersection  with  the  post  at 
B,  because  the  point  B is  perpendicular  to  P,  which  is  just  1 foot  from 
A,  for  the  post  batters  one  foot  in  rising  6 feet,  and  we  find  the  length 
of  the  brace  from  the  brace  table,  page  140,  to  be  8 feet  5 Tb0 o inches. 

If  the  distance  A A'  be  either  more  or  less  than  7 feet,  say  9 feet,  then 
the  line  A'P  may  be  found  by  proportion,  thus: 

7 : 6 ::  9 : the  required  distance  A'P. 

This  example  is  here  wrought  out  as  follows: 

9 7:6::  9:  7.714 

J 

7)54 

7.714 

32 

8.568  Am.  7 feet  8 jVo  inches. 

As  this  answer  is  a fractional  number,  the  length  of  the  brace  cannot 
be  found  from  the  brace  table,  but  must  be  obtained  by  calculation,  as  is 
explained  in  the  introduction  to  that  table,  by  reducing  the  number  to 
inches,  squaring  it  and  doubling  the  square,  and  then  extracting  the 
square  root  from  the  sum,  thus:  7 feet  8Td<fo  inches  = 92.56  inches,  and 
92.56  X 92.56  = 8567.3536,  double  this  and  we  have  the  following 
number,  17134.7072,  from  which  we  extract  the  square  root,  thus: 

17134.7072(130.89  Aw*., 

1_ 

23)71  or 

69 

2608)234.70  10  feet  lOyVo  inches, 

208.04 

26.0672  true  length  of  brace. 

23.5521 

2.5151 

The  bevels  of  the  long  upper  braces  will  be  the  same  as  those  of  the 
shorter  braces  already  found ; i.  e.,  all  the  upper  end  bevels  marked  B 
and  B'  are  equal  to  each  other,  and  all  the  lower  end  bevels  marked  A 
and  A'  arc  equal  ; — —the  bevels  and  lengths  of  the  braces  being  measured 
on  the  dotted  linesextending  from  toe  to  toe ; the  slant  heights  of  the 
intersections  of  the  braces  with  the  posts  at  A'  and  B'  are  found  as 
already  explained,  by  the  use  of  the  common  rafter  table,  or  by  the  u^e 
of  the  square. 


PLATE  41 


This  Plate  represents  an  improved  Straining-Beam  Bridge,  50  feet 
long,  for  a Railroad,  The  principles  of  its  construction  are  essentially 
the  same  as  those  of  the  bridges  represented  on  Plate  30 ; the  improve- 
ment of  having  the  straining  beams,  stringers,  and  main  braces,  each  in 
two  pieces  instead  of  one,  bolted  together,  yet  separated  by  packing  blocks, 
as  here  shown,  renders  them  much  stronger  and  more  durable ; and  the 
use  of  3 inch  planks,  10  inches  wide,  for  counter  braces,  and  the  lateral 
braces,  as  shown  in  the  floor-section,  effectually  meets  the  objections  to 
this  style  of  bridges,  as  stated  on  page  94 ; and  they  are  probably  as  in- 
expensive and  simple  in  construction  as  it  is  possible  for  a safe  bridge 
to  be  built. 

Fig.  1 shows  the  side  elevation,  and  Fig.  2,  the  floor-section,  without 
ties  or  rails,  to  exhibit  the  manner  of  framing.  The  counter  braces  are 
let  into  each  other,  at  their  intersection,  at  the  bottom,  1 inch  each,  and 
thus  locked  together;  the  supporting  rods  are  crossed,  as  recommended 
under  Plate  30.  Bills  of  timber  and  iron,  in  detail,  are  subjoined. 

In  the  State  of  Illinois,  the  cost  of  a bridge  of  this  kind  would  be 
about  $10  per  lineal  foot. 

Bill  of  Timber  for  One  Span  Improved  Straining  Beam  Bridge,  50  feet  long. 


2 Wall  plates, 

10  by 

12 

in., 

20  feet  long,  B.  M.  = 400  feet. 

8 Bolsters, 

10  by 

10 

a 

12 

n 

44 

“ = 800 

ft 

4 Stringers, 

7 by 

14 

a 

60 

tt 

it 

“ =l(i:« 

ft 

8 Main  braces, 

7 by 

10 

tt 

1G 

tt 

a 

“ = 7-17 

tt 

8 Minor  “ 

7 by 

8 

u 

14 

u 

tt 

“ = 52.! 

tt 

0 lateral  u 

6 by 

G 

u 

22 

tt 

tt 

“ = 

tt 

4 Straining  beams, 

7 by 

10 

it 

24 

a 

tt 

« = fit  id 

u 

25  Cross  floor  timbers, 

6 by 

13 

it 

18 

44 

tt 

“ =2025 

ti 

4 Kail  stringers, 

10  by 

12 

tt 

2G 

it 

a 

“ = 10  to 

it 

3 Nenlle  brains, 

10  by 

10 

ti 

2o 

it 

tt 

« = r,i  it  > 

tt 

4 ( oil  liter  Inaees, 

3 by 

10 

a 

18 

it 

tt 

« = ISO 

tt 

3 Parking, 

4 by 

i 1 

u 

12 

*1 

tt 

« = 5i> 

tt 

1 

(116) 

3 by 

10 

a 

] 2 

44 

it 

“ = :io 

SJ7DC 

ti 

nn 


BRIDGE  BUILDING, 


117 


Bill  c 

>f  Iron. 

8 

Supporting  rods, 

u 

in.  iron, 

12  feet, 

4 

in.  long, 

as=812 

lbs, 

4 

a u 

H 

44 

12  « 

4 

44 

sss210 

44 

4 

Cross  u 

1 

44 

17 

6 

44 

^178 

44 

8 

Brace  bolts, 

i 

44 

8 a 

6 

44 

« 80 

44 

18 

a « 

i 

44 

1 « 

8 

44 

= 45 

44 

4 

£ 

44 

1 « 

a 

46 

« 11 

44 

72 

lbs.  Nuts, 

■=  72 

44 

1358 

Bill  of 

Castings. 

16 

Washers, 

6 in. 

diam.  hole 

> 

1|  in. 

® 96 

lbs, 

8 

44 

6 

44  44 

1§  « 

= 48 

44 

60 

44 

• 

a 

46  m 

§ « 

= 75 

44 

8 

Lateral  angle  blocks, 

^160 

46 

379 


PLATE  42 


This  Plate  represents  the  manner  of  constructing  and  setting  a circu- 
lar ctnter  of  large  dimension,  for  the  purpose  of  springing  a heavy  stone 
arch.  Let  the  outline  of  the  frame  be  drawn  upon  a floor  or  platform, 
by  taking  OA  as  a radius,  as  described  on  page  67,  Plate  15.  It  will 
be  found  convenient  to  cut  a pattern  of  thin  stuff  for  the  12  pieces  of 
the  circular  rim,  by  which  those  pieces  are  to  be  made  of  2 inch  plank. 
One  of  them  should  be  cut  in  two  in  the  middle,  and  then  all  spiked 
together,  lapped  as  represented  in  the  plate.  The  3 middle  braces  ex- 
tend from  the  lower  edge  of  the  long  brace  to  the  upper  edge  of  the  rim, 
the  middle  braces  to  be  let  in  the  thickness  of  one  plank,  and  at  the 

lower  end  1 J inch  on  each  side,  and  spiked  to  the  rims  and  to  the  lower 

brace.  The  rims  should  not  be  more  than  30  inches  apart,  and  may  be 
Covered  with  good  fencing  laid  close  together. 

These  large  centers  should  be  set  to  their  places  by  wedges,  as  repre- 
sented in  the  plate,  or  they  could  not  easily  be  taken  out. 

Bill  of  Timber  for  One  Him  of  28  feet  span  and  10  feet  rise. 

12  pieces,  2 by  12  in.,  7 ft.  long. 

4 pieces,  2 by  10  in.,  14  ft.  long. 

3 braces,  6 by  6 in.,  12  ft.  long. 


(118) 


Sca  le  S feet  to  t/?e  inrA  . 


Plate  42. 


i 


The.o.  X.eorihardt  & Son.  PhiIa 


PLATE  43. 


This  P*ate  exhibits  a plain  and  excellent  plan  for  a frame-work  to 
support  a railroad  water-tank.  Fig.  1 represents  the  elevation  as  seen 
from  the  track,  and  Fig.  2 the  foundation  and  bottom  of  the  tank. 

The  proper  method  of  bracing  between  the  posts  is  shown  in  Fig.  1, 
and  the  dotted  lines  on  Fig.  2 show  the  proper  places  to  brace. 

The  workman  should  be  careful  to  have  the  braces  around  the  center 
all  run  the  same  way,  the  outer  braces  running  up  from  right  to  left 
and,  vice  versa,  the  inner  braces  running  up  from  left  to  right,  passing 
each  other  at  the  center ; both  series  to  be  framed  flush  with  the  posts 
on  their  respective  sides.  The  shoulders  and  bevels  of  all  the  braces 
must  be  framed  the  same  way,  so  that  changing  the  ends  of  a brace  will 
not  change  its  face  side.  The  braces  are  let  into  the  posts  not  less  than 
1 inch  to  give  support  to  the  toe,  and  the  posts,  in  this  example,  being  5 
feet  4 inches  apart,  the  run  of  the  braces  will  be  5 feet  6 inches,  and  the 
length,  on  a.  regular  run,  as  found  in  the  brace  table,  page  141 , is  7 feet 
9 rVa  inches.  Both  the  length  and  bevels  of  the  braces  are  measured  on 
the  dotted  lines  extending  from  toe  to  toe  in  Fig.  1,  the  bevels  being 
marked  at  45°  from  these  lines,  for  a regular  run.  These  explanations 
are  sufficient  to  enable  the  workman  to  construct  such  a husk  or  frame 
without  further  instruction.  Bills  of  timber  and  iron  are  given  in 
detail. 

Height  and  Location  of  Tank. 

The  height  from  the  top  of  the  rail  on  the  track  to  the  top  of  the 
floor  timbers  on  the  frame  work,  when  the  water  is  to  be  taken  out  of 
the  bottom  of  the  tank,  in  the  usual  way,  by  the  use  of  a goose-neck 
valve,  is  1 1 feet ; but  when  the  water  is  to  be  taken  from  the  side,  it 
should  be  about  1 foot  lower. 

For  a tank  of  22  feet  outside  diameter,  the  distance  of  the  center  of 
the  tank  from  the  middle  of  the  track  should  be  21  feet. 

To  Estimate  the  Capacity  of  a Tank. 

Carpenters  are  frequently  asked  to  give  the  capacity  of  a tank  or 
cistern,  which  it  is  difficult  for  them  to  do  without  some  rule  or  data, 
and  we  will  therefore  give  the  capacity  of  this  tank  and  the  rule  for 

(119) 


120 


CARPENTRY  MADE  EASY. 


obtaining  it.  The  inside  dimensions,  after  deducting  the  thickness  of 
the  staves,  bottom  and  chime,  are  as  follows : 


Top  diameter, 
Bottom  diameter, 
Mean  diameter, 
Height, 


20  ft.  8 in. 

21  ft.  6 in. 
21  ft.  2 in. 
15  ft.  4 in. 


To  obtain  the  mean  diameter  add  | of  the  difference  between  the  top 
and  bottom  diameters  to  the  top  diameter ; and  to  find  the  capacity,  multi- 
ply the  square  of  the  mean  diameter,  in  inches,  by  the  height,  in  inches, 
and  this  product  by  .0034 : the  answer  will  be  in  gallons.  This  decimal 
.0034  is  obtained  by  dividing  the  decimal  .7854  by  231,  the  number  of 
cubic  inches  in  a gallon ; and  the  decimal  .7854  is  used  to  reduce  the 
capacity  of  a square  tank  to  that  of  a circular  one,  for  .7854  is  the  area 
of  a circle  whose  diameter  is  one. 

To  make  it  more  plain  we  give  the  figures  in  full-^ 

Mean  diameter  = 254  in. 

Height  =184  in. 

254  x 254  x 184  X .0034  = 40361.20  gallons. 

Operation .«  254 

254 

1016 

1270 

508 

64516 

184 

258064 

516128 

64516 

11870944 

.0034 

47483776 

35612832 

40361.2096  Am* 

To  reduce  gallons  to  barrels  divide  by  31 J. 

To  find  the  circumference  of  a tank,  if  we  know  the  diameter,  we 
multiply  the  diameter  by  3.1416,  or  if  we  know  the  circumference  we 
can  find  the  diameter  by  dividing  it  by  3.1416 ; because  this  is  the  cir« 


RAILROAD  WATER-TANK. 


121 


cumferenee  of  a circle  whose  diameter  is  1.  Suppose  we  want  the  length 
of  the  bottom  hoop  on  a tank  22  feet  across  the  bottom— 

Operation .*  3.1416 

22 

62832 

62832 

69.1152  Ans,  69  ft.  1.38  in. 

12 

1.3824 


To  find  the  capacity  of  a square  cistern,  multiply  its  three  dimensions 
together,  viz.,  its  length,  breadth  and  height  in  inches,  and  divide  the 
product  by  231,  and  the  answer  will  be  in  gallons. 


Bill  of  Timber  for  Plate  43. 


2 Caps, 

12  by  12 

in. 

24 

feet  long,  Board 

measure 

= 576  feet 

2 Caps, 

12  by  12 

(t 

18 

a 

n 

<i 

ii 

= 432 

ii 

12  Posts, 

12  by  12 

« 

12 

<i 

a 

ii 

ii 

=1728 

ii 

6 Floor  timbers  6 by  12 

a 

24 

a 

n 

ii 

ii 

864 

(6 

2 “ “ 

6 by  12 

a 

22 

a 

a 

ii 

ii 

= 264 

a 

2 “ “ 

6 by  12 

n 

20 

a 

a 

ii 

ii 

= 240 

a 

2 “ “ 

6 by  12 

a 

18 

a 

<1 

ii 

ii 

= 216 

a 

2 a a 

6 by  12 

« 

16 

ii 

a 

ii 

ii 

= 192 

ii 

2 “ “ 

6 by  12 

« 

12 

a 

u 

ii 

ii 

= 144 

ii 

16  Braces, 

6 by  6 

u 

8 

n 

n 

ii 

ii 

~ 384 

n 

20  Rib  pieces, 

8 by  6 

a 

16 

a 

u 

ii 

u 

«=  480 

ii 

5520 

Bill  of  Iron  for  Same. 


16  Rods,  7 ft.  8 in.  long,  inch  iron, 
32  Cast  Washers,  2 lbs.  each, 

32  Nuts,  1 lb.  each, 


weight,  324  ll>s. 
64  lbs. 
_32  lbs. 

420  lbs. 


PLATE  44. 


Tins  Plate  illustrates  the  proper  manner  of  framing  and  constructing 
pile  bridges.  This  is  a difficult  and  perplexing  job  to  inexperienced 
workmen,  because  the  piles  cannot  be  driven  exactly  in  line  nor  at  ex- 
actly equal  distances  apart,  and  the  frame  must,  therefore,  be  fitted  to 
them.  After  the  piles  are  driven,  we  first  saw  them  off  to  the  proper 
height  for  the  top  of  the  tenons,  and  then  level  a bat-board  to  the  height 
of  toe  shoulder  and  nail  it  to  one  of  the  outside  piles,  as  represented  in 
Fig.  1 ; then  level  the  other  outside  pile,  in  the  same  manner,  and  lay  a 
straight  edge  upon  them  on  each  side  of  the  bent;  and  nail  bat-boards, 
by  it,  upon  both  sides  of  each  pile.  Then  strike  a chalk  line  on  the 
tops  of  the  piles,  and  lay  out  our  tenons ; then  place  the  square  upon  the 
bat-boards  and  mark  the  sides  of  the  tenons. 

Fig.  2 shows  one  bent  of  piles  cut  to  the  top  of  the  tenons  and  ready 
for  sawing  the  shoulders.  To  do  this,  place  the  saw  flat  upon  the  bat- 
boards,  saw  the  shoulders,  and  dress  off  the  tenons,  sizing  their  edges 
3 inches  thick  and  as  wide  as  they  will  square,  then  lay  a board  upon 
the  shoulders  of  the  piles,  along  the  whole  width  of  the  bent,  and  mark 
carefully  at  each  edge  of  the  tenons;  this  board  applied  to  the  cap  will 
give  the  proper  places  for  the  tenons,  their  lengths  and  their  distances 
apart.  The  draw-bores  should  be  1J  inches  in  diameter,  and  the  centers 
2J  inches  from  the  edge  of  the  tenon  and  1 J inches  from  the  shoulder. 
After  the  cap  has  been  faced  to  the  proper  size,  mark  the  draw-bores 
upon  it  2J  inches  from  the  ends  of  the  mortices  and  2 inches  from  the 
face,  which  gives  a draw  of  £ inch. 

By  framing  each  bent  in  this  manner,  the  caps  will  rest  square  upon 
the  shoulders  of  the  piles  and  make  a perfect  fit,  securing  both  greater 
strength  and  greater  durability  than  can  be  done  in  any  other  manner; 
for  it  is  not  uncommon  to  see  these  bridges  framed  so  loosely  that  the 
whole  hand  can  be  inserted  between  the  shoulder  and  the  cap. 

Fig.  3 represents  one  bent  of  a pile  bridge  finished ; with  lateral 
braces  bolted  to  the  piles  to  prevent  side  vibration  and  to  give  greater 
stability  to  the  work. 

Fig.  4 shows  the  side  elevation,  the  ends  of  the  ties  and  the  manner 
of  framing  the  guard-rail.  The  ties  are  5 by  8 inches,  except  every  4th 
tie,  which  should  be  6 by  8 inches,  and  let  down  over  the  stringer  with 
(122) 


BRIDGE  BUILDING. 


123 


a shoulder  of  1 inch  on  each  side,  to  prevent  the  track  from  slipping  on 
the  bridge,  and  on  curves  the  stringers  should  be  secured  to  the  caps  by 
long  spikes. 

The  guard-rail  should  be  § by  6 inches  and  cut  out  2 inches  over  each 
tie,  and  fastened  by  wrought-iron  spikes  driven  through  guard-rail  and 
tie  into  the  stringer.  The  guard-rail  is  used  for  the  double  purpose  of 
strengthening  the  bridge  and  preventing  the  train  from  getting  off  the 
bridge  should  it  jump  the  track.  The  ties  are  placed  6 inches  apart,  or 
14  inches  from  center  to  center. 

Fig.  5 represents  the  floor  section  without  either  ties  or  guard-rails, 
in  order  to  show  the  manner  of  securing  the  ends  of  the  rail-stringers. 
This  is  done  by  inserting  a 3 inch  oak  plank  about  3 feet  long,  as  rep- 
resented in  the  plate,  secured  by  f inch  bolts  9 inches  from  the  ends  of 
the  stringers,  and  3 inches  from  the  top  and  the  bottom,  on  each  aide  of 
the  joint,  making  8 bolts  to  each  bent. 


EXPLANATION  OF  THE  TABLES. 


Inanitions  of  Terms  and  Phrases  used  in  this  Explanation,  and  in  other 

I Fees  in  this  Work. 

The  LENGTH  of  a rafter  is  understood  to  be  measured  from  the  ex* 
fcreme  point  of  the  foot  to  the  extreme  point  r f its  upper  end.*  Hut 
in  these  Tables  no  allowances  are  made  for  the  projection  of  rafters  be- 
yond the  plate,  or  for  ridge  poles;  so  that  the  lenijth  of  common  rafters 
is  understood  to  be  the  distance  from  the  upper  and  outer  corner  of  the  plate 
to  the  very  peak  of  the  roof 

The  HUN  of  a rafter  is  the  horizontal  distance  from  the  extreme 
point  of  the  foot  to  a perpendicular  let  fail  from  the  upper  end.  In 
common  roofs y the  run  of  the  rafters  is  half  the  width  of  the  building. 

The  RISE  of  a rafter  is  the  perpendicular  distance  from  the  upper 
end  of  the  rafter  to  the  level  of  the  foot. 

The  GAIN  of  a rafter  is  the  difference  between  its  run  and  its  length. 
For  example,  a rafter  whose  run  is  12  feet,  and  whose  length  is  13 
feet,  has  1 foot  gain. 

The  learner  will  easily  perceive  that  the  length  of  any  rafter  is  the 
hypotenuse  of  a right-angled  triangle,  of  which  its  run  and  its  rise  are 
the  other  two  sides.  Hie  length  is  therefore  ascertained  with  perfect 
accuracy  by  adding  the  square  of  the  run  to  the  square  of  the  rise, 
and  extracting  the  square  root  of  their  sum.  (See  Part  1.,  Prop. 
XXIV.) 

Example  1.  The  length  of  a common  rafter  is  required  in  a build- 
ing 21  feet  wide,  the  roof  of  which  is  desired  to  have  a pitch  of  5 
indies  to  the  foot.  The  run  is  therefore  12  feet,  the  square  of  which 


* Except  in  hip  rafters,  the  length  of  which  is  always  to  be  measured  on  the  hacking , 
or  along  the  middle  line  of  the  upper  surface;  for  when  the  side  Level  is  all  cut  ot:  one 
side  of  the  upper  end,  as  it  sometimes  is,  then  the  point  of  the  rafter  will  extend  Lalf  it* 
thickness  beyond  its  estimated  length,  as  given  in  the  table,  &c. 


(127) 


128 


CARPENTRY  MADE  EASY. 


is  the  product  of  12  multiplied  by  12,  or  144  feet.  The  rise  is  12 
times  5 inches,  or  60  inches,  or  5 feet,  the  square  of  which  is  25 
feet ; which,  added  to  144,  makes  160  feet,  of  which  we  extract  the 
square  root  thus : . . 

(the  rule  for  which  may  be  found  in  any  common-  1)169(13 

school  arithmetic)  and  find  it  to  be  13  feet,  the  exact  \ 

length  of  the  rafter  required.  23)69 

00 

But  in  most  cases  the  result  is  obtained  in  the  form  of  a fraction 
and  it  will  be  found  convenient  to  reduce  the  run  and  the  rise  to  inches 
in  tiie  first  place,  and  then  the  root  is  obtained  in  inches  and  decimals 
of  an  inch,  which  can  be  carried  out  to  any  degree  of  accuracy  re- 
quired.  T.n  these  Tables  they  are  carried  to  hundredths  of  an  inch. 

Example  2.  Required  the  length  of  a rafter  for  the  building  de« 
scribed  in  Plate  4 of  this  work.  Width  of  building,  12  feet;  rise  of 
rafter,  6 inches  to  the  foot. 

The  run  of  the  rafter  is  6 feet,  or  72  in.,  of  which  the  square  is  5184 

“ “ “ « 1296 

6480 


rise 


or  06 


and  their  sum  is 

of  which  we  proceed  to  extract  the  square  root  thus: 
and  find  it  to  be  80  inches  and  49  hundredths  of  an 
inch  ; or,  6 feet  8 inches  and  ,V(),  as  given  in  the 
Table,  which  is  the  exact  length  of  the  required  rafter. 


8)6480(80.49 

64 


1604)8000 

6416 

16089)158400 

144801 


13599 


TABLE  I. 


TL^  r ge  ct  this  Table  is  to  furnish  the  practical  carpenter  with  the 
precise  lengths  of  common  rafters  for  buildings  of  all  sizes,  and  for 
roofs  of  every  pitch.  The  Table  is  carried  out  to  buildings  of  60  feet 
in  width ; but  should  the  length  of  rafters  for  a wider  building  be 
required,  it  will  be  necessary  to  add  such  numbers  together  in  the  left- 
hand  column  as  will  make  their  sum  equal  to  the  width  of  the  build- 
ing, and  then  the  sum  of  the  lengths  of  the  rafters  given  in  the  Table 
opposite  these  numbers,  thus  added  together,  will  be  the  true  length 
of  the  raders  required. 

For  example,  suppose  it  were  required  to  find  the  length  of  rafters 
for  a building  84  feet  wide,  6 inches  rise.  We  find  the  length  of  a 
rafter  of  half  that  width,  of  the  same  pitch,  to  be  28  ft.  5.74  in.,  and 
double  this  number  would  be  the  length  of  the  rafter  required,  or 
46  ft.  11.48  in. 

Example  2.  Kequired  the  length  of  the  rafters  for  a building  102 
feet  wide,  5 inches  rise.  W e perceive  that  50  and  52  added  together 
will  make  102.  The  lengths  of  these  two  dimensions  for  this  pitch, 
as  given  in  the  Table,  are  27  ft.  1 in.,  and  28  ft.  2 in.,  the  sum  of 
which  is  55  ft.  3 in.,  the  length  of  the  rafters  required. 

9 


(129) 


130 


CARPENTRY  MADE  EASY. 


TABLE  I. 

Length  of  Rafters  m Feet,  Inches,  and  Hundredths  of  an  Inch.  Rise  of  Raftes 

to  the  Foot  in  Inches. 


Js|  1 
§1*1 

O*.  J 

| 

1 inch 
Rise. 

2 inch 
Rise. 

8 inch 
Rise 

4 inch 
Rise. 

& inch 
Rise. 

6 inch 
Rise. 

7 inch 
Rise. 

8 inch 

Rise. 

f toefc  1 

31  w j 

1 

4 

2 1 

2 : 0.08 

2 : 0.33 

2: 

0.73 

2 : 

: 1.29 

2:  2.00 

2: 

2.83 

2 : 

3.78 

2: 

4. 84 

2 

6 

5 

2:  6j 

2 : 6.10 

2 : 6.41 

2: 

6.92 

2 : 

; 7.62 

2:  8.50 

2: 

9.54 

2: 

10.73 

3: 

0.05 

3 ; 

1.50 

6 

3 | 

3 : 0.12 

3 : 0.49 

3: 

1.10 

3: 

; 1.94 

3:  3 

3 : 

4.24 

3: 

5.67 

3: 

7.26 

3: 

9 

7 

3:  6 

3:  6.14 

3 : 6.57 

3: 

7.30 

3: 

: 8.27 

3:  9.50 

3 : 

10.95 

4 : 

0.62 

4: 

2.47 

4 : 

•1/10 

8 1 

4 

4:  0.16 

4 : 0.66 

4: 

1.47 

i: 

: 2.59 

4:  4 

4: 

5.66 

4: 

7.56 

4 : 

9.68 

5 : 

9 

4:  6 

4:  6.18 

4 : 6.74 

4: 

7.66 

4: 

: 8.92 

4 : 10.50 

5: 

0.36 

5 : 

2-51 

5 : 

4.89 

5 : 

7.50 

10 

5 

5 : 0.20 

5 : 0.82 

5 : 

1.84 

5: 

: 3.24 

5:  5 

5: 

7.08 

5 : 

: 9.45 

6 : 

0.10 

6: 

3 

11 

6:  6 

6 : 6.22 

5 : 6.91 

5: 

8.03 

5: 

; 9.57 

5 : 11.50 

6: 

1.78 

6: 

4.41 

6 : 

7.31 

6 : 

10.50 

12 

6 | 

6 : 0.25 

6 : 0.99 

6 : 

2.21 

6: 

; 3.89 

6:  6 

6: 

8.49 

6 : 

; 11.34 

7 : 

2.52 

7 : 

6 

11 

7 

7 : 0.29 

7 : 1.15 

7 : 

2.58 

7 : 

; 4.54 

7:  7 

7 : 

9.91 

8: 

1.23 

8 : 

5.94 

8 : 

9 

16 

8 

8:  0.33 

8 : 1.32 

8: 

2.95 

8: 

: 5.19 

8:  8 

8: 

11.32 

9 : 

: 3.12 

9 : 

7.36 

10  : 

18 

9 

9 : 0.36 

9 : 1.48 

9 : 

3.32 

9 ; 

: 5.84 

9:  9 

10: 

0.74 

10  : 

: 5.03 

10  : 

9.79 

11  : 

3 

20 

10 

10  : 0.40 

10  : 1,64 

10  : 

3.69 

10  ; 

: 6.49 

10  : 10 

11 : 

2.16 

11  : 

: 6.92 

12: 

0.22 

12: 

6 

22 

11 

11  : 0.44 

11  : 1.81 

11  : 

4.06 

11  : 

: 7.14 

11  : 11 

12  : 

3.58 

12 : 

: 8.81 

13  : 

2.64 

13  : 

9 

24 

12 

12 : 0.49 

12:  1.98 

12: 

4.43 

12 

: 7.79 

13: 

13  : 

4.99 

13  : 

: 10.70 

14  : 

5.06 

15  : 

26 

13 

13  : 0.53 

13  : 2.15 

13: 

4.80 

13 

: 8.44 

14:  1 

14: 

6.41 

15  : 

: 0.59  15: 

7.48 

16 : 

3 

28 

14 

14  : 0.57 

14:  2.31 

14: 

5.17 

I14 

: 9.09 

15:  2 

15 : 

7.82 

16  : 

: 2.48  16: 

9.90 

17  : 

6 

30 

15 

15  : 0.62 

15  : 2.48 

15: 

5.54 

15 

: 9.73 

16:  3 

16  : 

9.24 

17  : 

: 4.38 

18  : 

0.32 

18  : 

9 

32 

16 

16  : 0.66 

16:  2.64 

16 : 

5.91 

16 

: 10.38 

17:  4 

17  : 

10.65 

18  : 

: 6.27 

19  : 

2.74 

20  : 

34 

17 

17  : 0.70 

; 17  : 2.80 

17  : 

6.28 

17 

: 11.03 

18:  5 

19  : 

0.07 

19 

; 8.16 

20  : 

6.16 

21  : 

3 

36 

18 

18:  0.73 

18  : 2.96 

18: 

6.65 

18 

: 11.68 

19:  6 

20  : 

; 1.48 

20  ; 

: 10.06 

21  : 

7.58 

22: 

6 

38 

19 

19  : 0 78 

19 : 3.12 

19  : 

7.02 

20 

: 0.33 

20:  7 

21 : 

; 2.90 

21 

: 11.95 

22: 

10 

23- 

9 

40 

20 

20  : 0.81 

20  : 3.28 

20  : 

7.39 

21 

: 0.98 

21  : 8 

22: 

: 4.32 

23 

: 1.85 

24  : 

0.43 

35  : 

42 

21 

| 

21  : 0.85 

21  : 3.45 

21  : 

7.76 

22 

: 1.63 

22:  9 

23  ; 

: 5.74 

24: 

: 3.74 

25: 

2.85 

26  : 

3 

44 

22 

22 : 0.89 

22:  3.62 

22  : 

8.12 

23 

: 2.28 

23  : 10 

24: 

; 7.16 

25 

: 5.64 

26: 

; 5.27 

27  : 

: 6 

46 

23 

23  : 0.93 

23  : 3.79 

23 : 

8.49 

24 

: 2.93 

24:  11 

25  : 

: 8.58 

26 

: 7.53 

27  : 

: 7.69 

28  : 

; 9 

48 

24 

24  : 0.97 

24 : 3.96 

24: 

8.86 

25 

: 3.57 

26: 

26  : 

: 10 

27 

: 9.42 

28: 

: 10.12 

30 : 

60 

25 

25  : 1.01 

25  : 4.13 

25  : 

9.23 

26 

: 4.22 

27  : 1 

27  ; 

: 11.41 

28 

: 11.31 

30  : 

: 0.54 

31 : 

: 3 

62 

26 

1 26  : 1.06 

26  : 4.30 

26: 

9.60 

27 

: 4.87 

28:  2 

29  : 

: 0.83 

30 

: 1.20 

31  : 

; 2.96 

32  : 

: 6 

64 

27 

27  : 1.10 

27  : 4.46 

27  : 

9.97 

28 

: 5.52 

29:  3 

30  : 

: 2.24 

31 

: 3.09 

32: 

5.39 

S3 : 

: 9 

| 66 

28 

28:  1.14 

28  : 4.63 

28  : 

10.34 

29 

: 6.17 

30:  4 

31  : 

: 3.66 

32 

: 4.98 

33: 

; 7.81 

35: 

j 68 

29 

| 29  : 1.18 

29  : 4.79 

29  : 

10.71 

30 

: 6.82 

31  : 5 

32 

: 5.08 

33 

: 6.87 

34: 

: 10.24 

36: 

3 

L« 

30 

1 30  : 1.23 

30  : 4.96 

30  : 

11.08 

31 

: 7.47 

32:  6 

33 

: 6.49 

34 

: 8.76 

36: 

; 0.66 

1 37 

: 6 - 

EXPLANATION  OF  TABLES, 


131 


TABLE  I . — Continued. 

Length  of  Bafters  in  Feet,  Inches,  and  Hundredths  of  an  Inch.  R se  of  Raitev 

to  the  Foot  in  Inches. 


► 15 

Ran 

of 

Rafter. 

10  inch 
Rise. 

11  inch 
Rise. 

12  inch 

Rise. 

13  inch 
Rise. 

14  inch 
Rise. 

18  inch 
Rise. 

16  inch 
Rise. 

17  inch 
Rise. 

18  inch 
Rise. 

4 

2 : 

2:  7.24 

2:  8.55 

2:  9.94 

2 : 11.38 

2:  0.87 

3 : 2.41 

3:  4 

3:  5.61 

3:  7.26 

5 

2:  6 

3:  3.05 

3:  4.69 

3:  6.42 

3-  8.22 

3 : 10.09 

4 : 0.02 

4:  2 

4:  4.02 

4:  6.08 

6 

3 1 

3 : 10.86 

4:  0.83 

4:  2.91 

4:  5.07 

4:  7.31 

4 : 9.62 

5 : 

5:  2.42 

5 : 4 89 

7 

3:6 

4:  6.67 

4:  8.97 

4 : 11.39 

5:  1.92 

5:  4.53 

5 : 7.23 

5 : 10 

6:  0.82 

6:  3.71 

8 

4 1 

5:  2.48 

5:  5.11 

5:  7.88 

5 : 10.76 

6:  1.75 

6 : 4.83 

6:  8 

6 : 11.23 

7 : 2.53 

9 

4:  6 

5 : 10.29 

6:  1.25 

6 : 4.36 

6:  7.61 

6 : 10.97 

7 : 2.44 

7:  6 

7 : 9.63 

8:  1.34 

10 

5 

6 : 6.10 

6:  9.39 

7 : 0.85 

7:  4.45 

7 : 8.19 

8 : 0.04 

8:  4 

8:  8.04 

9:  0.16 

11 

5:  6 

7 : 1.91 

7 : 5.53 

7:  9.33 

8:  1.30 

1 8:  5.41 

8 : 9.65 

9:  2 

9:  6.44 

9 : 10.98 

12 

6 

7:  9.72 

8:  1.67 

8:  5.82 

8 : 10.15 

9:  2.63 

9 : 7.25 

10  : 

10:  4.84 

10:  9.79 

14 

7 

9 : 1.34 

9 : 5.95 

9 : 10.79 

10:  3.84 

10  : 9.07 

11  : 2.46 

11 : 8 

12:  1.65 

12:  7.43 

16 

8 

10:  4.96 

10  : 10.23 

11  : 3.76 

11  : 9.53 

12:  3.51 

12  : 9.67 

13:  4 

13  : 10.46 

14:  5.06 

18 

9 

11  : 8.58 

12:  2.51 

12:  8.73 

13:  3.22 

13:  9.95 

14  : 4.88 

15  : 

15:  7.27 

1 16  : 2.69 

20 

10 

13:  0.20 

13:  6.79 

14:  1.70 

14:  8.91 

15:  4.39 

16  : 0.09 

16:  8 

17:  4.08  18:  0.33 

22 

11 

14:  3.82 

14 : 11.07 

15:  6.67 

16:  2.60 

16  : 10.83 

17  : 7.30 

18:  4 

19  : 0.88 

] 19:  9.96 

24 

12 

jl5:  7.44 

16:  3.34 

16  : 11.64 

17  : 8.30 

18:  5.27 

19  : 2.51 

20: 

20:  9.69 

21  : 7.59 

26 

13 

16  : 11.06 

17:  7.62 

18:  4.61 

19  : 1.99 

19  : 11.71 

20  : 9.72 

21  : 8 

22:  6.50  j 23:  5.22 

28 

14 

18:  2.68 

18  : 11.90 

19  : 9.58 

20:  7.67 

21  : 6.14 

22  : 4.93 

23:  4 

24:  3.31 

25:  2.86 

30 

15 

19:  6.30 

20:  4.18 

21  : 2.55 

22  : 1.37 

23:  0.58 

24  : 0.14 

25: 

26:  0.12 

j 27  : 0.49 

32 

16 

20:  9.92 

21  : 8.46 

22:  7.52 

23:  7.07 

24:  7.02 

25  : 7.35 

26:  8 

27:  8.92 

28  : 10.12 

34 

17 

22:  1.54 

23:  0.74 

24:  0.49 

25 : 0.77 

26:  1.46 

27  : 2.56 

28:  4 

29:  5.73 

30:  7.76 

36 

18 

23:  5.16 

i 

24:  5.02 

25:  5.47 

26:  6.46 

27:  7.90 

28  : 9.77 

30: 

31  : 2.55 

32:  5.39 

38 

19 

24:  8.78 

25:  9.30 

26  : 10.44 

28:  0.15 

29:  2.34 

30  : 4.98 

31:  8 

32  : 11.36  34  : 3.02 

40 

20 

26:  0.40 

27:  1.57 

28:  3.41 

29  : 5.84 

30  : 8.70 

32  : 0.19 

33:  4 

34  : 8.17  1 36  : 0.66 

42 

21  | 

27:  4.02 

28:  5.85 

29  : 8.38 

30 : 11.53 

32:  3.22 

33  : 7.39 

35  : 

36:  4.97 

37  : 10.29 

44 

22 

28:  7.64 

29  : 10.13 

[si  : 1.35 

32:  5.22 

33:  9.66 

35  : 2.60 

36:  8 

38:  1.78 

39  : 7.92 

46 

23 

29  : 11.26 

31:  2.41 

32:  6.32 

33  : 10.92 

35:  4.10 

36  : 9.81 

38  : 4 

39  : 10.59 

41  : 5.55 

48 

21  | 

31  : 2.88 

32:  6.69 

33  : 11.29 

35:  4.61 

36  : 10.54 

38  : 5.02 

40  : 

41  : 7.40 

43:  3.19 

50 

25  | 

32:  6.50 

33  : 10.97 

35:  4.26 

36  : 10.30 

38:  4.98 

40  : 0.23 

41:  8 43:  4.21 

45:  0.82 

52 

26 

33  : 10.12 

35  : 3.25 

36:  9.23 

38:  3.99 

39  : 11.42 

41  : 7.44 

43:  4 45  : 1.01 

46  : 10.45  j 

54 

27 

35:  1.74 

36:  7.53 

38:  2.20 

39  : 9.68 

41  : 5.86 

43  : 2.65 

45  : 

46:  9.82 

48:  8.09  j 

56 

23  | 

36:  5.36 

37  : 11.80 

39:  7.17 

41  : 3.37 

43:  0.30 

44  : 9.86 

46:  8 48:  6.63 

50.  5.72  | 

58 

29  | 

37:  8.98 

39:  4.08 

41  : 0.14 

42:  9.06 

44:  6.74 

46  : 5.07 

48:  4 50:  3.44 

52:  3.35  j 

60 

30  1 

39:  0.60 

i40:  8.36 

42:  5.11 

44:  2.75 

46:  1.18 

48  : 0.28 

50  : 

|52:  0.25 

54: 

TABLE  II. 


Length  of  Hip  Rafters. 

If  a roof  were  perfectly  horizontal  or  flat,  the  hip  rafters  would  each 
be  equal  to  the  diagonal  of  a square,  having  for  its  side  half  the  width 
of  the  building;  and  the  square  root  of  twice  the  square  of  half  the 
side,  would,  in  that  case,  be  the  length  of  the  hip  rafter.  This  we  call 
the  RUN  of  the  hip  rafter.  But  if  the  roof  has  any  pitch,  the  length  of 
the  rafter  is  greater  than  its  run , and  is  always  equal  to  the  hypotenuse 
of  a right-angled  triangle,  having  the  run  for  a base,  and  the  rise  for  a 
perpendicular;  and  the  length  is  found,  as  in  common  rafters,  by  add- 
ing the  square  of  the  run  to  the  square  of  the  rise,  and  extracting  the 
square  root  of  the  sum. 

Two  calculations  are  necessary  according  to  the  above  demonstra- 
tion : First,  for  obtaining  the  run  ; and  secondly,  having  found  the  run, 
from  that  to  obtain  the  length. 

First.  Suppose  the  width  of  the  building  to  be  40  feet,  and  the  rise 
of  the  roof  5 inches  to  the  foot,  or  100  inches ; then  half  the  width  of 
the  building,  20  feet,  is  240  inches,  the  square  of  which  is  57,600. 
Double  this  number  (for  the  two  sides  of  the  square)  is  115,200  inches, 
of  which  the  square  root  is  339  jV©  inches,  or,  28  ft.  3T40©  in.,  which 
is  the  run  of  the  hip  rafter . 

Second.  To  obtain  the  length , which  equals  the  hypotenuse  of  a 
right-angled  triangle,  of  which  the  run  is  the  base  and  the  rise  the 
perpendicular. 

The  run  is  339.41  inches  as  obtained  above, 

The  square  of  which  is  115,200 

The  rise  is  100  inches,  of  which  the  square  is  10,000 

The  sum  of  these  two  squares  is  125,200  inches,  of  which  the  square 
root  is  353.83  inches,  or  29  ft.  5.83  in.,  which  is  the  true  length  of  the 
hip  rafter. 

The  process  of  obtaining  the  length  is  explained  above  according  to 
the  long  way , and  the  most  obvious  and  analytical  way  also,  and  one 
which  every  practical  mechanic  should  make  himself  fully  familiar 
with  : but  practically , the  process  may  be  shortened  as  follows: — 
(132) 


EXPLANATION  OF  TABLES. 


133 


Add  the  square  of  the  rise  to  twice  the  square  of  half  the  widths  and  the 
square  root  of  the  sum  will  he  length  of  hip  rafter  required.  Thus : — - 
The  square  of  the  rise  (100  inches)  is  10,000 

Twice  the  square  of  half  the  width  is  115,200 

Their  sum  is  125,200 

of  which  the  square  root  is  353.83  in.,  or  29  ft.  5.83  in.  as  before 
which  is  the  true  length  of  the  hip  rafter  as  measured  on  the  backing. 
See  note  on  p.  127. 

TABLE  II. 


Length  of  Hip  Baiters  in  Feet,  Inches,  and  Hundredths  of  an  Inch. 


Sfl 

a 

U 

« 

li 

nch 

3 inch 

3 inch 

4 inch 

6 inch 

6 inch 

7 inch 

8 inch 

IIs 

c<3 

(4 

Bise. 

Bise. 

Bise. 

Bise. 

Bise. 

Bise. 

Bise. 

Bise. 

4 

2: 

9.94 

2: 

10 

2: 

10.17 

2: 

10.46 

2: 

10.87 

2: 

11.38 

3: 

3: 

0.71 

3: 

1.52 

5 

3: 

6.42 

3: 

6.50 

3: 

6.72 

3: 

7.08 

3: 

7.58 

3: 

8.22 

3: 

9 

3: 

9.89 

3: 

10.90 

6 

4: 

2.91 

4: 

3 

4: 

3.26 

4: 

3.70 

4: 

4.30 

4: 

5.07 

4: 

6 

4: 

7.07 

4: 

8.28 

7 

4: 

11  39 

4: 

11.50 

4: 

11.80 

5: 

0.31 

5: 

1.02 

5: 

1.92 

5 : 

3 

5 : 

4.25 

5: 

5.66 

8 

5: 

7.88 

5 : 

8 

5: 

8.35 

5 : 

8.93 

5: 

9.74 

5; 

10.76 

6: 

6: 

1.43 

6: 

3.04 

9 

6: 

4.36 

6: 

4.50 

6: 

4.89 

6: 

5.55 

6: 

6.46 

6: 

7.61 

6: 

9 

6: 

10.60 

7: 

0.42 

10 

7: 

0.85 

7: 

1 

7: 

1.44 

7: 

2.16 

7: 

3.17 

7: 

4.45 

7: 

6 

7: 

7.78 

7: 

9.80 

11 

7: 

9.33 

7 : 

9,50 

7: 

9.98 

7: 

10.78 

7: 

11.89 

8: 

1.30 

8: 

3 

8: 

4.96 

8 : 

7.18 

12 

8: 

5.82 

8: 

6 

8: 

6.52 

8: 

7.39 

8: 

8.61 

8: 

10.15 

9 : 

9: 

2.14 

9: 

4.66 

14 

9: 

10.79 

9: 

11 

9: 

11.61 

10: 

0.63 

10: 

2.04 

10: 

3.84 

10 : 

6 

10: 

8.50 

10: 

11.33 

16 

11 : 

3.76 

11 : 

4 

11: 

4.70 

11 : 

5.86 

11 : 

7.48 

11: 

9.53 

12: 

12: 

2.86 

12: 

6.09 

18 

12: 

8.73 

12: 

9 

12: 

9.79 

12: 

11.09 

13: 

0.91 

13: 

3.22 

13: 

6 

13: 

9.21 

14: 

0.85 

20 

14: 

1.70 

14: 

2 

14: 

2.88 

14: 

4.33 

14: 

6.35 

14: 

8.91 

15  : 

15: 

3.57 

15: 

7.61 

22 

15 : 

6.67 

15: 

7 

15: 

7.96 

15: 

9.57 

15 : 

11.79 

16: 

2.60 

16: 

6 

16: 

9.93 

17: 

2.37 

24 

16: 

11.64 

! 17 : 

17  : 

1.05 

17: 

2.80 

17: 

5.22 

17: 

8.30 

18: 

18: 

4.29 

18: 

9.34 

26 

18: 

4.61 

18  : 

5 

18: 

6.14 

18: 

8.03 

18: 

10.66 

19: 

1.99 

19: 

6 

19: 

10.64 

20: 

3.90 

28 

19 : 

9.58 

19  : 

10 

19: 

11,23 

20: 

1.26 

20: 

4.09 

20: 

7.68 

21  : 

21 : 

5 

21 : 

10.66 

30 

21 : 

2.55 

21  : 

3 

21 : 

4.32 

21 : 

6.50 

21: 

9.53 

22: 

1.37 

22: 

6 

22: 

11.36 

23: 

5.42 

82 

22: 

7.52 

22: 

8 

22: 

9.40 

22: 

11.73 

23: 

2.96 

23: 

7.06 

24: 

24: 

5.72 

24: 

11.18 

31 

24  : 

0.49 

24: 

1 

24: 

2.49 

24: 

4.97 

24: 

8.40 

25: 

0.76 

25: 

6 

26: 

0.07 

26: 

5.94 

36 

25: 

5.46 

25: 

6 

25: 

7.58 

25: 

10.20 

26: 

1.83 

26: 

6.45 

27: 

27: 

6.43 

28: 

0.70 

38 

26: 

10.44 

26  : 

11 

27  : 

0.67 

27: 

3.43 

27: 

7.27 

28: 

0.14 

28: 

6 

29  : 

0.79 

29: 

7.47 

40 

1 33: 

3.4^ 

28: 

4 

28: 

5.76 

28: 

8.27 

29: 

0.71 

29: 

6.83 

30: 

30: 

7.15 

31: 

2.23 

TABLE  III. 


Hip  and  Jack  Rafters  on  Octagonal  Roofs 

The  length  of  one  side  of  an  octagonal  building  being  commonly 
given  as  the  basis  of  calculation  in  framing,  it  will  first  be  necessary 
from  this  basis,  to  determine  with  accuracy  the  width  of  the  building 
from  the  middle  of  one  side  to  the  middle  of  the  opposite  side ; and 
also  the  diagonal  width,  from  one  corner  to  the  opposite  corner. 

The  width  FGr  (in  Plate  20,  Fig.  1)  is  obviously  the  same  as  one 
side  of  the  circumscribed  square  DE ; and  DE  is  made  up  of  three 
parts,  namely,  DA,  AB,  and  BE,  one  of  which  parts,  AB,  is  known- 
being  a side  of  the  given  octagon.  The  other  two  parts  are  equal  to 
each  other,  namely,  DA=BE  * We  have,  therefore,  to  find  the  length 
of  DA,  to  double  it,  and  to  add  AB  to  it  in  order  to  ascertain  the 
width  of  the  building.  The  length  of  DA  is  found  as  follows : — 

In  the  right-angled  triangle  CAD,  the  hypotenuse  AC,  being  one 
of  the  sides  of  the  given  octagon,  is  known ; and  the  square  of  this 
hypotenuse  is  equal  to  the  sum  of  the  squares  of  the  other  two  sides 
DA  and  DC,  or  to  double  the  square  of  DA. 

For  example,  suppose  the  sides  of  the  regular  octagon  be  given 
equal  to  16  feet,  which,  on  reducing  it  to  inches  to  insure  greater  accu- 
racy, is  192  inches. 

The  square  of  192  inches  is  36,864  inches,  one  half  which  is  18,482 
inches,  which  is  the  square  of  DA. 

The  square  root  of  18,432  inches  is  135.76  inches,  or  11  ft.  3.76  in., 
the  length  of  DA. 

Double  this  number  (for  DAxBE),  and  add  16  feet  for  the  length 
of  AB,  and  we  have  33  ft.  7.52  in.,  the  width  of  the  building. 

The  diagonal  width  is  obtained  as  follows : 

Let  O'  represent  the  point  at  the  foot  of  the  perpendicular  let  fall 


• The  equality  of  DA  and  BE  may  be  demonstrated  thus: — Suppose  the  figure  di- 
vided into  two  parts  by  the  line  FG,  and  these  two  parts  to  be  folded  together,  the  line 
FG  forming  the  fold;  then  the  point  A would  fall  upon  the  point  B,  the  point  C upon  the 
point  H,  and  the  point  D upon  the  point  E ; otherwise,  the  polygon  is  not  a regular  poly- 
gon, nor  the  circumscribed  square  a perfect  square.  In  a similar  manner,  it  may  be 
demonstrated  that  the  line  AD  is  equal  to  DC,  by  supposing  the  figure  to  be  folded  upon 
the  line  DL. 

(134, 


EXPLANATION  OF  TABLES. 


135 


from  O,  tlie  apex  of  the  roof,  upon  the  plane  or  level  of  the  plates  CA, 
AB,  &c. 

Then,  in  the  right-angled  triangle  O'FA,  the  sum  of  the  squares  of 
the  two  sides  AF  and  FO'  will  equal  the  square  of  the  hypotenuse 
AO' 

Having  found  above  that  FG=38  ft.  7.52  in.,  then  FO'  will 
equal  half  this  number,  or  19  ft.  3.76  in.,  or  231.76  inches,  the 
gquare  of  which  is  53,712,6976  inches. 

FA  is  half  of  the  given  side  AB,  and  is  8 feet,  or  96  inches,  of  which 
the  square  is  9,216  inches,  which,  being  added  to  53,712.6976  inches, 
is  62,928.6976  inches,  the  square  root  of  which  is  250.85  inches,  or 
20  ft.  10.85  in.,  the  length  of  AO',  or  half  the  diagonal  width  of  the 
building.  Double  this  number  is  41  ft.  9.70  in.,  the  length  of  APt 
the  diagonal  width  required . 

Half  the  diagonal  width  is  of  course  the  run  of  the  hip  rafters , and  half 
the  square  width  is  the  run  of  the  middle  jack  rafters  ; and,  having  ascer- 
tained these,  the  lengths  of  the  rafters  are  calculated  according  to  the 
rule  given  at  the  commencement  of  this  general  explanation  of  the 
Tables — by  taking  the  square  root  of  the  sum  of  the  squares  of  the  run 
and  the  rise  of  any  given  rafter. 


II.  Width  of  Octagon  given  to  find  the  Diagonal  and  the  Side. 

It  sometimes  happens,  as  in  church  spires  for  example,  that  the 
width  of  an  octagon  is  given,  from  which  the  other  dimensions  must 
be  found. 

Let  PC,  the  width  of  a regular  octagon,  P E 

be  given,  to  find  AB  the  side,  and  AE 
the  diagonal. 

Draw  OD  from  the  centre  of  the  octa- 
gon to  an  angle  of  the  circumscribed 
square.  Then  OD2=CD2-f  OC2,  or  20C2, 
since  OD  is  the  hypotenuse  of  the  tri- 
angle ODC,  of  which  the  other  two  sides 
OC  and  DC  are  equal  to  each  other,  and 
each  one  equals  half  the  given  width  of 
the  octagon.  Then,  since  OA  bisects  the  vertical  angle  of  the  triangle 
COD,  it  divides  the  base  into  two  segments,  which  are  proportional  to 
the  adjacent  sides  (Part  I.,  Prop.  XXVI.);  and  we  have  the  following 
proportion : 


DO  : OC : : DA  : AC ; 


136 


CARPENTRY  MADE  EASY. 


and,  by  composition, 

DO+OC  : OC  : : DA+AC  : AC; 
but  AC  is  half  the  required  side  AB. 

Having  obtained,  by  the  above  formula,  the  length  of  AC,  it  will 
be  easy  to  obtain  that  of  OA,  since  OA9=OC2+AC2. 

Example.  Suppose  PC,  the  given  width,  equals  10  ft.  8 in.,  which 
is  the  width  of  the  base  of  the  church  spire  described  in  Plate 
28.  Reducing  this  number  to  inches,  to  insure  greater  accuracy,  we 
have  PC =128  inches.  And  OD  would  then  equal  the  square  root  of 
the  sum  of  the  squares  of  OC  and  DC,  each  of  which  equals  64  inches. 
Double  the  square  of  64  inches  equals  8192  inches,  the  square  root  of 
which  is  90.51  inches,  which  is  the  length  of  OD ; then,  by  applying 
or  substituting  this  value  in  the  first  proportion  given  above,  we  have 
90.51  in.:  64  in.::  DA:  AC; 

and  DA  being  yet  unknown,  we  ascertain  it  by  composition,  thus : 
90.51+64  : 64 : : DA+AC,  or  DC : AC ; 
or,  154.51 : 64 : : 64  : AC. 

Multiplying  the  middle  terms  of  this  proportion  together,  and  divid- 
ing the  product  by  the  first  term,  we  have  the  value  of  the  last  term, 
or  AC,  equal  to  26.51  inches.  Double  this,  and  we  have  AB,  the  re- 
quired side , equal  to  53.02  inches,  or  4 ft.  5.02  in.,  and  OC2+AC2= 
AO2,  or  702.78+4096=4798.78  inches,  the  square  root  of  which  is 
69.27  inches,  or  5 ft.  9.27  in. ; and  the  whole  of  the  required  diagonal 
equals  twice  this  number,  or  11  ft.  6.54  in. 

Note. — Since  all  regular  octagons  are  similar  figures,  any  two  regular  octagons  of  dif- 
ferent dimensions  will  not  only  have  their  sides  proportional,  but  their  widths  and  their 
diagonal  widths  proportional  also ; and  if  we  have  the  exact  dimensions  of  all  the  parts 
of  one  octagon  given,  and  any  one  part  of  the  other  octagon  also  given,  then  all  its  re- 
maining parts  can  be  found  by  proportion. 

Example  1.  Required  the  diagonal  width  of  a regular  octagon,  the 
aide  of  which  is  12  feet. 

Let  us  compare  this  with  another  octagon,  all  the  dimensions  of 
which  we  know,  or  which  we  can  find  from  the  Table ; say,  an  octagon 
of  16  feet  side,  the  diagonal  width  of  which,  as  given  in  the  Table,  is 


EXPLANATION  OF  TABLES.  137 

41  ft.  9.7  in  ; tnen,  since  all  the  parts  of  the  one  figure  are  propoi* 
tional  to  the  corresponding  parts  of  the  other,  we  shall  have 
side  to  side,  as  diagonal  width  to  diagonal  width ; 
or,  16  : 12  : : 41  9.7  to  the  answer  =31  ft.  4.27  in., 

Thus : 

We  multiply  the  second  and  third  terms  together,  and  divide  by  the 
first: 

41  ft.  9.7  in. 

12 

492 

9.7 

16)501.7(31.356  feet  ana  decimals  of  a foot,  which  we  reduce  to 
48  feet  and  inches,  thus : 


21 

31.356 

16 

12 

”57 

i27 

48 

~90 

80 

Too 

By  multiplying  the  tenths  of  a foot  by  12  to  bring  them  to  inches, 
and  disregarding  the  third  decimal  figure,  we  have  for  a final  answer 
31  ft.  4t2075  in.  as  the  required  answer ; which  is  verified  by  the  num- 
ber as  given  in  the  Table. 

Example  2.  Required  the  side  of  a regular  octagon,  the  square  width 
of  which  is  30  feet.  We  compare  this  with  the  same  octagon  as  be- 
fore, and  have 

width  to  width  as  side  to  side ; 

or,  38  7.52  : 30  : : 16 : to  the  answer  =12  feet  5.12  inches, 

Thus: 

as  in  this  case  the  first  term  or  divisor  is  a compound  number,  ire 
reduce  all  the  three  given  terms  to  inches,  and  have 
38  ft.  7.52  in.  =463.52  inches, 

30  feet  =360  inches, 

16  feet  =192  inches. 


138  CARPENTRY  MADE  EASY. 

So  that  the  proportion,  in  inches,  is 

463.52  : 360  : : 192  : answer  =12  fit.  5.12  in. 
360 


11520 

576 


463.52)69120(149.119  in.,  or  12  ft.  5.12  in.,  Am. 

46352 


227680 

185408 


422720 

417168 

55520 

46352 


91680 

46352 


€58280 


EXPLANATION  OF  TABLES. 


139 


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TABLE  IT 

The  length  of  braces , like  the  length  of  ratters,  n*ast  be  determined 
by  extracting  the  square  root  of  the  sum  of  the  squares  of  the  perpen 
dicular  and  the  horizontal  runs. 

This  table  embraces  almost  every  length  that  can  be  required  in 
framing  buildings,  and  comprises  those  of  both  regular  and  irregular 
runs. 

The  exact  length  is  here  given  to  the  hundredth  part  of  an  inch; 
but  practically  it  will  be  found  best  to  cut  each  brace  from  a sixteenth 
an  eighth  of  an  inch  longer  than  the  exact  rule  requires,  in  ordei 
compensate  for  compression  and  the  shrinkage  of  the  timber. 

(140) 


EXPLANATION  OF  TABLES.  141 

TABLE  IV. 

•Length  of  Braces  given  in  Feet,  Inches,  and  Hundredths  of  an  Inch. 


Length  of  Bun. 

Length  of  Brace. 

1 Length  of  Bun. 

Length  of  Brace. 

Length  of  Run. 

Length  of  Brae*. 

ft.  la.  Ft.  la  . 

Ft.  In.  Ft.  In. 

Ft.  In.  Ft.  In. 

0 ■ 6 by  0 : 6 

0:  8.48 

3 : 3 by  3 : 3 

4:  7.15 

5 : 6 by  5:6 

7:  9.33 

0 : 6 by  0 : 9 

0 : 10.81 

3 : 3 by  3 : 6 

4:  9.31 

5 : 9 by  5:9 

8:  1.58 

0 : 9 by  0 : 9 

1:  0.72 

3 : 3 by  3 : 9 

4 : 11.54 

6 : 0 by  6:0 

8:  5.82 

0 : 9 by  1 : 0 

Is  3 

3 : 3 by  4 : 0 

5:  1.84 

6 : 3 by  6:3 

8 : 10.06 

1 : 0 by  1 : 0 

1:  4.97 

3 : 6 by  3 : 6 

4 : 11.39 

6 : 6 by  6:6 

9:  2.30 

1 : 0 by  1 : 3 

1:  7.20 

3 : 6 by  3 : 9 

A : 1.55 

6 : 9 by  6:9 

9:  6.55 

1 : 3 by  1 : 3 

1 : 9.23 

3 : 6 by  4 : 0 

6:  3.78 

7 : 0 by  7:0 

9 : 10.79 

1 : 3 by  1 : 6 

1 : 11.43 

3 : 9 by  3 : 9 

5:  3.63 

7 : 3 by  7:3 

10:  3.03 

1 : 6 by  1 : 6 

2 : 1.45 

3 : 9 by  4 : 0 

5:  5.79 

7 : 6 by  7:6 

10:  7 28 

1 : 6 by  1 : 9 

2:  3.65 

4 : 0 by  4 : 5 

5:  7.88 

7 : 9 by  / : 9 

10 : 11.52 

1 : 9 by  1 : 9 

2:  5.69 

4 : 0 by  4 : 3 

5 : 10.03 

8 • & by  8:0 

11  : 3.76 

1 : 9 by  2 : 0 

2:  7.89 

4 : 0 by  4 : 6 | 

[ fi  * o 25 

8 : 3 by  8:3 

11:  8 

2 : 0 by  2 : 0 

2:  9.94 

4 : 0 by  4 : 9 

6:  2.51 

8 : 6 by  8:6 

12:  0.24 

2 : 0 by  2 : 3 

3:  0.12 

4 : 0 by  5 : 0 

6 : 4.83 

8 : 9 by  8:9 

12:  4.49 

2 : 0 by  2 : 6 

3:  2.41 

4 : 3 by  4 : 3 

6:  0.12 

9 : 0 by  9:0 

12:  8.73 

2 : 3 by  2 : 6 

3:  4.36 

4 : 3 by  4 : 6 

6:  2.27 

9 : 6 by  9:6 

13:  5.22 

2 : 6 by  2 : 6 

3:  6.42 

4 : 3 by  4 : 9 

6:  4.49 

10  : 0 by  10  : 0 

14:  1.70 

2 : 6 by  2 : 9 

3:  8.59 

4 : 3 by  5 : 0 

6 : 6 74 

10  : 6 by  10  : 6 

14  : 10.19 

2 : 9 by  2 : 9 

3 : 10.66 

4 : 6 by  4 : 6 

6:  4.36 

11  : 0 by  11  : 0 

15:  6.67 

2 : 9 by  3 : 0 

4:  0.83 

1 4 : 6 by  4 : 9 

6:  6.51 

11  : 6 by  11 : 6 

10:  3.16 

3 : 0 by  3 : 0 

4:  2.91 

4 : 6 by  5 : 0 

6:  8.72 

12  : 0 by  12  : 0 

J6: 11.64 

3 : 0 by  3 : 3 

4:  5.02 

4 : 9 by  4 : 9 

6:  8.61 

12  : 6 by  12  : 6 

17:  8.13 

3 . 0 by  3 : 6 

4:  7.31 

4 : 9 by  5 : 0 

6 : 10.75 

13  : 0 by  13  : 0 

18:  4.61 

3 ; 0 by  3 : 9 

4:  9.62 

5 : 0 by  5 : 0 

7 : 0.85 

13  : 6 by  13  : 6 

19:  1.10 

3 : 0 by  4 : 0 | 5 : 

5 : 3 by  5 : 3 

7:  5.09 

1 14:0  by  14:0 

19:  9.58 

142  CARPENTRY  MADE  EASY. 

TABLE  V. 

Weight  of  Square  Iron  in  Pounds  and  Ounces. 


Size. 

1 foot. 

S feet. 

8 feet.  | 

4 feet. 

S feet. 

6 feet. 

1 feet. 

8 feet,  j 

9 feet. 

Inches. 

lbs. 

ox. 

lbs.  oz. 

lbs. 

OZ. 

lbs. 

OZ* 

lbs. 

OZ. 

lbs. 

OZ. 

lbs. 

oz. 

lbs.  < 

OZ. 

lbs.  i 

0ft, 

* 

0: 

13 

1 : 11 

2: 

8 

2: 

6 

4: 

3 

5: 

2 

5: 

15 

6: 

13 

7 : 

10 

\ % 

1: 

5 

2: 

10 

4 : 

0 

5: 

5 

6 : 

10 

7 : 

9 

9 : 

5 

10  : 

10 

12: 

0 

X 

l: 

15 

3: 

14 

5 : 

13 

7 : 

12 

9 : 

10 

11  : 

9 

13: 

7 

15: 

6 

17: 

4 

X 

2: 

10 

5 : 

4 

7 : 

14 

10: 

8 

13  : 

0 

15 : 

8 

18  : 

2 

20 : 

12 

23: 

6 

1 

3: 

6 

6 : 

12 

10: 

2 

13: 

5 

16  : 

12 

20: 

4 

23  : 

10 

27  : 

0 

30: 

6 

IX 

5 

8 : 

10 

12  : 

15 

17  : 

3 

21  : 

8 

25  : 

12 

30 : 

0 

34  : 

5 

38: 

9 

ix 

5: 

5 

10  : 

10 

15: 

15 

21 : 

2 

26  : 

7 

31  : 

12 

37  : 

0 

42  : 

4 

47: 

8 

1% 

6 : 

6 

12  : 

12 

19  : 

3 

25: 

9 

32: 

0 1 

38: 

6 

44: 

12 

51 : 

2 

57  : 

•1 

ix 

7: 

10 

15  : 

4 

22: 

12 

30  : 

6 

38  : 

0 ! 

45  : 

10 

53  : 

*\ 

60: 

14 

68  : 

7 j 

ix 

8: 

14 

17  : 

12 

26  : 

11 

35  : 

9 

44  : 

8 ! 

53  : 

7 

62: 

6 

71  : 

5 

80: 

4 1 

ix 

10: 

6 

20  : 

12 

31 : 

2 

41  : 

7 

51  : 

13 

62: 

2 

72: 

8 

82: 

14 

93: 

4 

IX  I 

11  : 

14 

23 

12 

35  : 

10 

47  : 

8 

59  : 

6 | 

71  : 

5 

83  : 

3 

95: 

2 

107  : 

0 

2 1 

13  : 

8 

27 

0 

40  : 

10 

54  : 

2 

67  : 

10  | 

81  : 

2 

94: 

10 

102: 

03 

121  : 

11 

2X 

15 : 

5 

30  : 

8 

45  : 

13 

61  : 

2 

76  : 

6 

91  : 

10  | 

106  : 

13 

122: 

2 

137  : 

6 

2X 

17  : 

2 

34  : 

3 

51  : 

5 

68  : 

6 

85  : 

9 

102  : 

14 

119  : 

13 

136  : 

15 

154: 

o i 

2x  ! 

19: 

1 

38: 

2 

57  : 

8 

76: 

4 

95: 

. 

5 1 

114  : 

6 

133  : 

7 

152: 

8 

171 : 

1 

9 

2X 

21  : 

1 

42  : 

3 

63  : 

; 6 

84  : 

: 8 

105  : 

10  1 

126  : 

14 

147  : 

15 

169  : 

0 

190 : 

: 1 

2X  1 

j 23  : 

5 

46  : 

10 

69  : 

: 15 

93  : 

: 3 

116  : 

; 8 

139  : 

: 13 

163 : 

1 

186  : 

5 

209  : 

; 10 

2X 

25  : 

10 

51  : 

: 02 

76  : 

12 

102  : 

: 3 

I 127  : 

: 13 

153: 

; 6 

178  : 

: 15 

204  : 

: 8 

| 230: 

: 0 

2X  | 

27  : 

14 

55  : 

: 14 

83  : 

: 13 

111  : 

: 12 

139  : 

: 11 

; 167  : 

: 10 

195  : 

9 

1 223 : 

: 8 

251  : 

: 7 

3 ' 

j 30  : 

6 

60  : 

: 12 

91  : 

: 3 

121  : 

: 9 

152  : 

: 2 

! 182 : 

: 8 

212  : 

: 14 

243  : 

: 5 

273: 

: 11 

3'X  i 

33  : 

: 0 

66  : 

: 0 

99  : 

: 0 

1 132 : 

: 0 

165  : 1 

! 198  : 

1 

231  : 

: 1 

264  : 

: 2 

297  : 

: 2 

3X  1 

35  : 

: 11 

71 

: 6 

107  : 

: 2 

j 142 : 

: 13 

I 178 : 

: 8 

i 214 

: 3 

249 

: 15 

1 285 

: 10 

321 

: 5 

3X  i 

38  : 

: 8 

77  : 

: 0 

115  : 

: 8 

154 

: 0 

192 

: 8 

231 

: 0 

269 

: 8 

308 

: 0 

j 346 

: 8 

8X 

41  : 

: 6 

82: 

: 12 

124  : 

: 3 

167 

: 9 

j 207 

: 0 

248 

: 6 

289 

: 12 

331 

: 3 

j 372 

: 10 

3X 

44  : 

: 6 

88 

: 12 

133 

: 4 

1 177 

: 11 

221 

: 2 

266 

: 8 

310 

: 14 

j 355 

: 5 

399 

: 13 

3X 

47 

: 8 

95 

: 1 

142 

: 9 

| 190 

: 2 

237 

: 11 

285 

: 3 

332 

: 11 

j 380 

: 5 

427 

: 13 

j 3* 

50 

: 13 

101 

: 8 

152 

: 5 

203 

: 0 

253 

: 13 

304 

: 8 

355 

: 5 

406 

: 0 

456 

: 13 

! 4 

54 

: 2 

108 

: 3 

162 

: 4 

216 

: 5 

270 

: 6 

324 

: 8 

376 

: 10 

432 

: 11 

486 

: 13 

4X 

57 

: 8 

115 

: 0 

172 

: 10 

230 

: 2 

1 287 

: 10 

1 345 

: 2 

402 

: 10 

460 

: 2 

517 

: 11 

4 X 

| 61 

: 1 

| 122 

: 2 

183 

: 3 

244 

: 4 

305 

: 5 

1 366 

: 6 

427 

•.  7 

488 

: 7 

649 

: 8 

4X 

64 

: 11 

| 129 

: 6 

194 

: 2 

258 

: 13 

! 323 

• 8 

j 388 

: 3 

452 

: 14 

517 

: 10 

582 

: 5 

4X 

68 

: 6 

136 

: 14 

205 

: 5 

273 

: 13 

342  : 3 

1 410 

: 11 

479 

: 2 

647 

: 10 

616 

: 0 

| 4Vt 

1 72 

: 5 

144 

: 10 

216 

: 15 

289  : 3 

I 361 

: 8 

1 433 

: 13 

506 

: 2 

578 

: 7 

45C 

: 11 

1 76 

: 5 

152 

: 8 

228 

: 12 

305 

: 1 

i 381 

: > 

4,57 

: 9 

533 

: 13 

610 

: 1 

686 

: 0 

|_2L 

80 

: 6 

160 

: 11 

241 

: 0 

221 

: 5 

401 

: 11 

482 

: 0 

562 

: 5 

642 

: 11 

723 

: 0 

MS_ 

Li4 

: 8 

| 169 

: 0 

253 

• 6 

337 

: 15' 

422 

: 6 

1 606 

: 15 

591 

: 6 

675 

. 14 

760 

: 6 

EXPLANATION  OF  TABLES. 
TABLE  V. — Continued. 


143 


Weight  of  Square  Iron  in  Pounds  and  Ounces. 


M feet. 

11  feet. 

12  feet. 

13  feet. 

14  feet. 

IS  feet. 

16  feet. 

17  feet. 

18  feet. 

Inches. 

lbs.  oz. 

lbs.  oz. 

lbz.  oz. 

lbs.  oz. 

lbs.  oz. 

lbs.  oz. 

lbs.  oz. 

lbs.  oz. 

lbs.  OL 

lA 

8:  8 

9:  5 

10:  2 

11 : 0 

11  : 13 

12:  9 

13:  8 | 

14:  6 

15:  3 

% 

13:  5 

14:  10 

16:  0 

17:  5 

18  : 10 

19  : 15 

21  : 4 

22:  9 

23  : 14 

v< 

19  : 0 

20  : 15 

22: 13 

24  : 12 

26  : 10 

28:  8 | 

30:  6 

32:  4 

34  : 02 

% 

28:  0 

8 

31:  2 

33  : 12 

36:  6 

38 : 14 

41 : 8 

44:  2 

46  : 12 

i 

33  : 12 

37:  3 

40  : 10 

44:  0 

47  : 6 

50  : 12 

54:  2 

67:  8 

60 : 14 

42  : 14 

47  : 2 

51  : 6 

55  : 11 

59  : 15 

64:  4 

68:  8 

72 : 13- 

77:  0 

52 : 12 

58:  1 

63:  6 

68  : 11 

74:  0 

79:  4 

84:  8 

89  : 13 

95:  0 

1 X 

63:  14 

70:  4 

76  : 10 

83:  0 

89:  7 

95  : 14 

102:  2 

108:  9 

115:  0 

76:  0 

83  : 10 

91  : 4 

98  : 14 

106:  8 

114:  2 

121  : 12 

129  : 6 

13?:  0 

j ik 

89:  4 

98:  3 

107:  2 

116:  1 

125:  0 

133  : 15 

142 : 14 

151  : 13 

160 : 11 

103  : 10 

114:  0 

124:  5 

134  : 11 

145:  0 

155  : 5 

165  : 11 

176  : 0 

186:  6 

118  : 13 

130  : 11 

142 : 10 

154:  8 

166:  6 

! 178 : 3 

ICO:  2 

202:  0 

2i  3 : 14 

2 

135:  3 

148  : 11 

162:  3 

175  : 12 

189:  4 

202  : 13 

216:  5 

229  : 13 

243  : 6 

2 % 

152  : 10 

167  : 14 

183:  3 

198:  7 

213  : 11 

229  : 0 

244:  5 

259  : 9 

274  : 14 

2 % 

171  : 2 

188:  3 

205:  5 

222:  8 

239  : 10 

256  : 11 

273  : 13 

290  : 15 

308:  0 

2% 

190  : 10 

209  : 11 

228  : 12 

247  : 14 

266  : 15 

286:  0 

305:  1 

324  : 2 

343:  3 

..  zy»  \ 

211  : 3 

232:  5 

253:  7 

274:  9 

295' : 12 

316  : 14 

337  : 15 

359:  0 

380:  3 

1 2% 

232  : 15 

256:  3 

279  : 8 

302 : 13 

326  : 2 

349  : 7 

372  : 12 

396:  0 

419  : 5 

2%  ! 

| 255  : 10 

281  : 3 

306  : 13 

332:  6 

357  : 13 

383  : 6 

409  : 

434:  8 

460:  1 

2^ 

279  : 6 

307  : 5 

335:  4 

363:  3 

391  : 1 

419:  0 

447:  0 

475:  0 

502:  14 

3 

301:  2 

334:  9 

365:  0 

395  : 6 

425  : 13 

456  : 3 

486  : 10 

517:  1 

547  : 8 

3* 

330:  2 

363  : 2 

396:  2 

329  : 2 

462:  2 

495:  3 

528:  3 

561  : 3 

594:  3 

3X 

357  : 0 

392  : 11 

428:  6 

464:  3 

500:  0 

535  : 11 

571:  5 

607:  0 

642 : 11 

3% 

385:  0 

423:  8 

462:  0 

500  : 8 

539  : 0 

577:  8 

616:  0 

654  : 9 

693:  1 

; 3 * 

414  : 1 

j 455  : 8 

496  : 15 

538:  5 

379  : 11 

621  : 2 

662:  8 

703  : 14 

745  : 5 

3% 

444  : 3 

488:  9 

533:  0 

| 

577  : 6 

621  : 15 

666:  5 

710 : 11 

755  : 2 

799:  8 • 

| 3%  | 

475:  5 

522 : 14 

570:  6 

617  : 15 

665:  8 

713:  0 

760:  8 

808:  1 

855 : 10 

3^  | 

507  : 10 

558:  5 

609  : 1 

659  : 13 

710  : 10 

761  : 5 

812:  2 

862  : 15 

913 : 10 

4 

540  : 14 

594  : 15 

649  : 0 

703  : 2 

757  : 3 

811:  4 

865  : 5 

919  : 6 

973:  8 

575:  3 

632  : 11 

690:  3 

747  : 11 

805:  3 

862  : 12 

920:  5 

977  : 13 

1035  : 5 ■ 

4^ 

610:  9 

671  : 10 

732  : 11 

793  : 12 

854  : 13 

915  : 14 

976 : 15 

1037  : 15 

1099  : 0 

4X 

646:  0 

711  : 11 

776:  6 

841  : 2 

905  : 13 

970:  8 

1035  : 3 

1099  : 14 

1164  : 10 

4>i 

6S4  : 8 

752  : 15 

821  : 6 

8S9  : 13 

958:  5 

1026 : 11 

1095:  3 

1163  : 10 

1232:  2 

4% 

721  : 1 

1 795  : 6 

867  : 11 

940:  0 

1012:  5 

1084  : 10 

1156 : 15 

1229  : 3 

1301:  8 

4% 

762  : 10 

838  : 15 

915  : 3 

1 991  : 7 

1067  : 11 

1144:  0 

1220:  3 

1296:  8 

1372 : 13 

V/t 

803:  5 

883  : 11 

964:  0 

1044:  5 

1124  : 11 

1205:  0 

1285  : 5 

1365  : 11 

1446.  0 

i ® 

1 644  : 13 

929  : 5 

1013:  14 

1098:  3 

1182 : 11 

12b7:  S 

1351  : 11 

1436:  3 

1 1520  ; 10 

144 


CARPENTRY  MADE  EASY. 


TABLE  YI. 

Weight  of  Plat  Iron  in  Pounds  and  Ounces. 


rTfctok. 

i foot. 

t feet. 

S feet.  | 

4 feet. 

* feet. 

• feet. 

7 feet. 

• feet. 

t feet. 

I*  feat. 

lasher 

Inches. 

lbs. 

or 

lbs. 

or 

lbs. 

os. 

lbs. 

OS* 

lbs. 

or 

Ibs. 

or 

lbs. 

oz. 

lbs. 

or 

lbs. 

or 

lbs. 

ea. 

' 

* 1 

0: 

13  | 

1 : 

11 

2: 

8 

3: 

6 

4: 

2 

5: 

0 

5: 

14 

6 : 

13 

7: 

10 

8: 

8 

1: 

1 

2: 

2 

3: 

3 

4 : 

3 

5 : 

4 

6: 

5 

7: 

6 

8: 

7 

9: 

8 

10: 

9 

1 : 

4 

2: 

8 

3 : 

12 

5 : 

1 

6: 

5 

7: 

9 

8: 

13 

10: 

1 

11 : 

5 

12: 

10 

1 : 

8 

3: 

0 

4: 

7 

5: 

14 

7: 

6 

8: 

14 

10: 

6 

11 : 

13 

13: 

5 

14  : 

13 

2 j 

1 : 

11 

3 : 

6 

5: 

1 

6 : 

13 

8: 

8 

10: 

1 

11 : 

33 

13: 

8 

15 : 

3 

16  : 

14 

2X 

1 : 

14 

3 : 

12 

5 : 

11 

7 : 

10 

9: 

8 

11 : 

6 

13  : 

4 

15  : 

3 

17  : 

1 

19  : 

0 

2 X 

2: 

1 

4 : 

3 

6: 

5 

8 : 

7 

10  : 

9 

12  : 

11 

14: 

13 

16  : 

14 

19  : 

0 

21 : 

1 

2X 

2: 

5 

4 : 

10 

7 : 

0 

9 : 

5 

11 : 

10 

13  : 

14 

16  : 

5 

18: 

10 

20: 

15 

22: 

: 3 

3 

2 : 

8 

5 : 

1 

7 : 

10 

10  : 

1 

12: 

11 

15  : 

3 

17  : 

11 

20: 

5 

22: 

13 

25: 

: 5 

2: 

12 

5 : 

8 

8: 

3 

11  : 

0 

13: 

12 

16  : 

8 

19  : 

3 

22: 

: 0 

24: 

12 

27  : 

: 8 

3 X 

3 : 

0 

5 

14 

8: 

13 

11  : 

12 

14: 

11 

17  : 

10 

20  : 

10 

23: 

: 9 

26: 

9 

29: 

; 8 

3 x ! 

3 : 

3 

6 

5 

9 : 

8 

12: 

11 

15  : 

13 

19  : 

0 

22: 

3 

25  ; 

: 6 

28: 

8 

31 

: 11 

« 

3: 

6 

6 

13 

10  : 

2 

13  : 

8 

16  : 

14 

20  : 

5 

23: 

11 

27: 

: 0 

30  : 

6 

33 

: 13 

1 

1 : 

4 

2: 

8 

3 : 

12 

5 : 

: 1 

6: 

5 

7 : 

9 

8 : 

: 14 

10 : 

: 2 

11  : 

6 

12 

: 11 

>*l 

1 : 

10 

3 : 

4 

4: 

13 

6: 

5 

7 : 

14 

9 : 

: 8 

11 : 

; 1 

12 

: 11 

14: 

5 

15 

: 13 

IX 

1 : 

14 

3 : 

12 

5 : 

11 

7 ; 

; 10 

9 : 

8 

11  : 

: 6 

13  : 

: 5 

15.: 

: 2 

17 : 

1 

19 

: 0 

IX 

2 : 

3 

4 : 

6 

6: 

10 

8 : 

; 14 

11 : 

1 

13  : 

: 5 

15 : 

: 8 

17: 

: 11 

20: 

1 

22 

: 3 

2 

2: 

8 

5 : 

; 1 

7 : 

9 

10  ; 

: 1 

12: 

9 

15  : 

: 2 

17  : 

; 11 

20 

: 5 

22: 

11 

25 

: 6 

2X 

2 : 

14 

5 : 

; 11 

8 : 

8 

11  ; 

; 6 

14: 

4 

17  : 

; 2 

20; 

: 0 

22 

: 12 

25  : 

; 10 

28 

: 8 

x- 

2X 

3 : 

3 

6 : 

; 6 

9 : 

8 

12 

: 11 

15: 

13 

19  : 

: 0 

22; 

: 3 

25 

: 6 

28: 

: 8 

31 

: 11 

2X 

3 : 

: 8 

7: 

: 0 

10  : 

8 

13  ; 

: 15 

17  : 

7 

20: 

: 14 

24: 

: 7 

27 

: 15 

31 : 

: 7 

31 

: 15 

3 

3: 

13 

7 : 

: 10 

11  : 

6 

15: 

: 3 

19  : 

; 0 

22 

: 12 

26: 

: 9 

30 

: 6 

34: 

: 3 

38 

: 0 

$x 

4 : 

2 

8 : 

: 4 

12: 

6 

16  : 

: 8 

20  ; 

; 10 

24 

: 12 

28 

: 14 

33 

: 0 

37  : 

: 2 

41 

: 4 

3X 

4: 

; 7 

8; 

: 14 

13: 

5 

17; 

: 12 

22; 

; 3 

26 

: 10 

31 

: 0 

35 

: 7 

39  ; 

: 15 

44 

: 6 

*X 

4 ; 

: 12 

9 : 

: 8 

14  : 

; 4 

19 

: 0 

23  : 

: 12 

28 

: 8 

33 

: 4 

38 

: 0 

42; 

: 12 

47 

: 8 

4 

5 : 

: 1 

10; 

: 2 

15: 

; 3 

20: 

: 4 

25; 

: 5 

30 

: 6 

35 

: 8 

40 

: 9 

45; 

: 10 

50 

: 11 

1 

1 ; 

: 11 

3 ; 

: 6 

5 : 

; 1 

6 

: 13 

8: 

: 8 

10 

: 2 

11 

: 13 

13 

: 8 

15: 

: 3 

16 

: 14 

IX 

2: 

; 1 

4 

: 3 

6 : 

; 6 

8 

: 7 

10  ; 

• 9 

12 

: 11 

14 

: 13 

16 

: 14 

19  : 

: 0 

21 

: 2 

ix 

2: 

: 8 

5 

: 1 

7 : 

; 9 

10 

: 1 

12; 

: 10 

15 

: 3 

17 

: 11 

20 

: 5 

22: 

: 13 

25 

: 5 

ix 

3 

: 0 

5 

: 15 

8; 

; 14 

11 

: 13 

14: 

: 12 

17 

: 11 

20 

: 10 

23 

: 1C 

26: 

: 9 

29 

: 9 

2 

3 

: 6 

6 

: 12 

10; 

2 

13 

: 8 

16  : 

: 14 

20 

: 5 

23 

: 10 

27 

: 0 

30  : 

: 6 

33 

: 12 

2X 

3 

: 13 

7 

: 10 

11  : 

; 7 

15 

: 3 

1 19: 

: 0 

22 

: 13 

26 

: 10 

30 

: 7 

34: 

: 4 

38 

: 0 

2X 

4 

: 3 

8 

: 6 

12; 

: 10 

16 

: 14 

21 

: 2 

35 

: 5 

26 

: 9 

¥ 

: 14 

38: 

: 2 

42 

: 4 

<M 

I 4 

: 10 

9 

: 5 

13  ; 

: 15 

18 

: 9 

23: 

: 3 

27 

: 14 

32 

: 8 

37 

: 3 

41  : 

: 13 

46 

: 8 

3 

|»:  1 

10 

: 2 

15  ; 

: 3 

20 

: 4 

25; 

: 5 

30 

: 6 

35 

: 8 

40 

: 9 

45  : 

: 10 

50 

: 11 

3* 

6 

: 8 

11 

: 0 

16  ; 

: 8 

22 

: 0 

27 

: 8 

33 

: 0 

38 

: 8 

44 

: 0 

49: 

: 8 

55 

: 0 

. 3* 

6 

: 14 

11 

: 12 

17 

: 11 

23 

: 10 

29 

: 9 

35 

: 8 

41 

: 7 

47 

: 5 

63  . 

4 

69 

: S 

explanation  of  tables.  145 

TABLE  VI. — Continued. 

Weight  of  Flat  Iron  in  Pounds  and  Ounces. 


Thick. 

Wide. 

1 foot. 

2 feet. 

3 feet.  | 

4 feet. 

6 feet. 

6 feet. 

7 feet. 

8 feet. 

9 feet. 

1C  feetH 

laches. 

Inches. 

lbs.  oz. 

lbs.  oz. 

lbs.  oz.  | 

lbs.  oz. 

lbs.  oz. 

lbs.  oz. 

lbs.  oz. 

lbs.  oz. 

lbs.  oz. 

lbs.  os. 

K. 

K- 

K- 

1 

3X 

6 : ’o 

12:  11 

19:  0 

25:  5 

31  : 11 

38:  0 

44:  6 

50  : 11 

57:  0 

63:  5 

4 

6 : 12 

13:  8 

20:  5 

27:  0 

33  : 12 

40:  9 

47:  4 

54:  0 

60:  12 

67  : 9 

1 

2:  1 

4:  3 

6:  5 

8:  7 

10  : 9 

12  : 11 

14  : 13 

16  : 15 

19:  0 

21:  2 

1>4 

2 : 10 

5:  5 

7 : 15 

10:  9 

13:  3 

15  : 13 

18:  8 

21:  2 

23:  12 

26:  6 

IX 

3:  3 

6:  6 

9:  8 

12:  11 

15  : 13 

19:  0 

22:  3 

25:  6 

28:  8 

31 : 11 

3:  11 

7:  6 

11  : 1 

14:  13 

18:  8 

22:  3 

25 : 14 

29:  9 

33:  4 

37:  0 

2 l 

4:  3 

8:  6 

12 : 10 

16:  14 

21  : 1 

25:  5 

29:  9 

33:  12 

38:  0 

42:  3 

2K 

4 : 12 

9:  8 

14:  5 

19:  0 

23  : 12 

28:  8 

33:  5 

38:  0 

42 : 12 

47:  8 

2K 

5 : 5 

10:  9 

15  : 13 

21  : 1 

26:  6 

31 : 10 

37:  0 

42:  3 

47  : 8 

52:  12 

2 X 

5 : 13 

11 : 10 

17:  6 

23:  3 

29:  0 

34 : 13 

40  : 11 

46:  8 

52:  5 

58:  2 

3 

6 : 5 

12: 11 

19:  0 

25  : 5 

31  : 11 

38:  0 

44:  6 

50  : 11 

57  : 10 

63  : 0 

3K 

6 : 14 

13  : 12 

20:  9 

27:  8 

34 : 5 

41  : 3 

48:  1 

55:  0 

61  : 12 

68:  10 

3K 

7:  6 

14  : 12 

22:  3 

29:  9 

37:  0 

44:  6 

51  : 13 

59:  3 

66:  8 

73:  14 

3*  i 

7 : 14 

15  : 12 

23 : 11 

31  : 10 

39:  9 

47:  8 

55:  7 

63:  6 

71:  5 

79:  4 

4 ||  8:  7 

16  : 14 

25:  5 

33:  12 

42:  3 

50  : 10 

59:  1 

67:  9 

76:  0 

84:  8 

2:  8 

5:  1 

7:  9 

10:  1 

12  : 10 

15  : 3 

17  : 11 

20:  5 

22  : 13 

25:  5 

IK 

3:  3 

6:  6 

9:  8 

12:  10 

15:  13 

19:  0 

22:  3 

25  : 6 

28:  8 

31  : 11 

IK 

3: 12 

7:  9 

11 : 6 

15:  3 

19:  0 

22:  12 

26:  9 

30:  6 

34:  3 

38:  0 

IK 

4:  7 

8:  14 

13:  5 

17  : 12 

22:  3 

26:  10 

31  : 1 

35:  8 

39  : 15 

44:  6 

2 

5:  1 

10:  2 

15:  3 

20  r 4 

25:  5 

30:  6 

35:  8 

40:  9 

45 : 10 

50:  11 

2K 

5:  11 

11 : 6 

17:  1 

22:  12 

28:  8 

34:  3 

39  : 15 

45  : 10 

51:  5 

57:  0 

2K 

6:  5 

12:  10 

19:  0 

25:  5 

31  : 10 

38:  0 

44:  6 

50  : 11 

57:  0 

63:  5 

2K 

7 : 0 

14:  0 

21:  0 

28:  0 

35:  0 

42:  0 

49:  0 

56:  0 

63:  0 

69:  0 

3 

7:  10 

15:  4 

22: 13 

30:  6 

38  : 0 

45  : 10 

53:  4 

60  : 14 

68:  8 

76:  0 

3K 

8:  4 

16 : 8 

24:  12 

33:  0 

41  : 4 

49:  7 

57  : 12 

66:  0 

74:  3 

82:  7 j 

3K 

8:  14 

17  : 12 

26  : 10 

35:  8 

44:  6 

53:  4 

62:  6 

71:  0 

79  : 14 

88  : 12 1 

3K 

9 : 8 

19:  0 

28  : 8 

38:  0 

47:  8 

57:  0 

66:  8 

76:  0 

85:  8 

95:  0 

4 

10:  2 

20:  4 

30:  6 

.40:  9 

50  : 11 

60  : 14 

70  : 15 

81:  1 

91  : 3 

101:  6 

IK 

5:  1 

10:  2 

15:  3 

20:  4 

25:  5 

30:  6 

35:  8 

40:  9 

45  : 10 

50  : 11 

J 

2 

6 : 12 

13:  8 

20:  4 

27:  0 

33:  12 

40:  9 

47:  5 

64:  0 

60:  12 

67:  9] 

3 

10:  2 

20:  4 

30:  6 

40:  9 

50  : 11 

60  : 13 

70:  15 

81  : 1 

91  : 3 

101:  5] 

4 

13:  8 

27:  0 

40:  9 

54:  1 

67  : 9 

81:  1 

94:  9 

108:  1 

121 : 10 

135:  2 

5 

10:  14 

33 : 12 

50  : 11 

67:  9 

84:  8 

101 : 6 

118:  5 

135:  4 

152:  2 

169:  0 

L 6 

20:  5 

40:  10 

60 : 15 

81  : 2 | 101 : 5 

121  : 11 

141  : 14!  162:  3 

182:  8 

202:12.1 

146  CARPENTRY  MADE  EASY. 

TABLE  VII. 

Weight  of  Round  Iron  in  Pounds  and  Ounces. 


Size. 

1 foot. 

S feet. 

s feet. 

4 feet. 

6 feet. 

« feet. 

7 feet. 

• feet.  | 

• feet. 

Diameter 

in  inches. 

lbs. 

os. 

lbs. 

OS. 

lbs. 

OI. 

lbs. 

OS. 

lbs. 

os. 

lbs.  , 

os. 

Ibs. 

os. 

lbs.  os.  | 

ibs.  « 

ML 

X 

0 : 

11 

1 : 

5 

2: 

0 

2 : 

11 

3: 

5 

4: 

0 

4: 

: 11 

5: 

5 i 

6: 

0 

X 

1: 

0 

2: 

1 

3: 

2 

4: 

3 

5 : 

4 

6: 

5 

7; 

: 5 

8 : 

6 I 

9: 

7 

* 

1 : 

8 

3: 

0 

4: 

8 

6: 

0 

7 : 

8 

9: 

0 

10: 

: 8 

12: 

0 

13  : 

8 

X 

2: 

0 

4: 

1 

6: 

1 

8 : 

2 

10: 

3 

12: 

3 

14 

: 4 

16  : 

5 

18: 

5 

l 

2: 

11 

6: 

5 

8: 

0 

10: 

10 

13  : 

5 

15  : 

15 

18: 

: 10 

21 : 

3 

23: 

11 

W 

3 : 

6 

6; 

12 

10: 

2 

13: 

7 

16  : 

13 

20 : 

3 

23 

: 8 

26: 

14 

30: 

4 

IX 

4: 

3 

8: 

6 

12: 

8 

16: 

11 

20: 

14 

25: 

0 

29 

: 3 

33: 

6 

37: 

8 

w 

6: 

0 

10  : 

0 

15 : 

1 

20: 

1 

25: 

2 

30 : 

2 

35 

: 2 

40: 

3 

45 : 

3 

ix 

6: 

0 

11 : 

15 

17: 

15 

23: 

14 

29  : 

14 

35: 

13 

41 

: 12 

47: 

12 

53 : 

11 

ix 

7 : 

0 

14  : 

0 

21: 

0 

28  : 

0 

35: 

1 

42: 

1 

49 

: 1 

56: 

1 

63: 

1 

ix 

8: 

2 

16: 

4 

24: 

6 

32: 

8 

40: 

10 

48: 

12 

56 

: 14 

65  : 

0 

72: 

3 

1% 

9: 

5 

18  : 

11 

28  : 

37  : 

5 

46  : 

: 11 

56: 

0 

65 

: 5 

74: 

11 

84: 

0 

2 

10 : 

10 

21 : 

4 

31 

14 

42: 

8 

53: 

2 

63  : 

12 

74 

: 5 

84: 

14 

95 : 

8 

) %x 

12 : 

24: 

36: 

48: 

60: 

72: 

84: 

96  : 

108: 

) *X 

13: 

8 

26  : 

14 

40  : 

6 

53  : 

13 

67  : 

4 

80: 

9 

94 

: 2 

107  : 

8 

121 : 

2% 

15: 

30: 

45 : 

60: 

75: 

89  : 

15 

104 

: 13 

119  : 

13 

134: 

13 

23  X 

16: 

11 

33: 

4 

50: 

1 

66: 

12 

83: 

7 

100: 

1 

116 

: 12 

133: 

8 

150: 

3 

2% 

18  : 

13 

37  : 

10 

54: 

6 

73  : 

3 

92: 

: 10 

112 : 

8 

131 

: 4 

150: 

0 

168: 

12 

2^ 

20: 

1 

40  : 

3 

60  : 

5 

80  : 

6 

10C  : 

: 7 

120: 

8 

140 

: 9 

160 : 

10 

180: 

11 

2^ 

21  : 

15 

43: 

14 

65  : 

13 

S7  : 

12 

109  : 

: 11 

131 : 

10 

153 

: 9 

175: 

8 

197: 

7 

3 

(23; 

14 

47  : 

12 

71  : 

11 

95: 

9 

119  : 

: 7 

143: 

5 

167 

: 3 

191 : 

1 

215: 

0 

3* 

25: 

14 

51  : 

13 

77  : 

12 

103  : 

11 

129: 

: 10 

155 : 

9 

181 

: 8 

207  : 

7 

233  : 

6 

s* 

28: 

0 

66: 

1 

84: 

2 

112: 

3 

140: 

: 3 

168: 

4 

196 

: 5 

224: 

5 

253  : 

6 

3X 

30: 

4 

60  : 

5 

90: 

12 

121 : 

0 

151 : 

: 4 

181 : 

7 

211 

: 11 

241 : 

15 

272: 

S 

3X 

32: 

8 

65 : 

0 

97  : 

8 

130  : 

0 

162  : 

: 9 

195  : 

1 

127 

: 9 

260  : 

1 

292: 

9 

3X 

34: 

14 

69 : 

12 

104  : 

10 

139  : 

8 

174: 

: 6 

209  : 

5 

244 

: 3 

279  : 

1 

314: 

0 

3X 

37  : 

5 

74  : 

11 

112: 

. 

0 

149  : 

5 

186  : 

: 11 

224: 

0 

261 

: 5 

298  : 

11 

336: 

0 

3 X 

39: 

14 

79  : 

12 

119  : 

10 

159: 

8 

199  : 

: 5 

239  : 

3 

279 

: 0 

318: 

14 

358: 

12 

4 

42: 

8 

84: 

15 

127  : 

6 

169  : 

14 

213  : 

: 5 

254  : 

13 

297 

: 4 

339  : 

12 

382: 

■ i 
4 , 

4X 

45; 

3 

90  : 

5 

135  : 

8 

180  : 

: 11 

225  • 

• 14 

271  : 

0 

316 

: 4 

361  : 

7 

406: 

10 

4* 

48  : 

0 

95 : 

15 

143  : 

14 

191  : 

13 

239  ; 

: 12 

287  : 

11 

335 

: 10 

383: 

9 

431 : 

9 

*X 

50: 

12 

| 101  : 

11 

152: 

7 

203: 

5 

254: 

: 1 

304  : 

15 

355 

: 11 

406: 

8 

457  : 

6 

*x 

53  : 

12 

107  : 

: 8 

161  : 

5 

215 : 

0 

268: 

: 12 

322  : 

11 

376 

: 5 

430: 

2 

483: 

13 

4X 

56  : 

12 

113  : 

10 

170  : 

: 6 

227  : 

3 

283  : 

: 15 

340  : 

11 

397 

: 8 

454: 

5 

511 : 

2 

4X 

60  : 

0 

119  : 

14 

179  : 

; 12 

239  : 

10 

299: 

: 8 

359  : 

6 

419 

: 6 

479  : 

2 

639 : 

1 

*X 

63: 

2 

126: 

3 

189  : 

; 6 

252  : 

: 6 

315: 

: 8 

378: 

10 

441 

: 11 

504  : 

12 

667  : 

13 

6 

1 66: 

12 

133  : 

; 8 

200: 

6 

267  : 

0 

333: 

: 12 

400: 

8 

467 

: 5 1 

534  : 

0 1 

600: 

“J 

EXPLANATION  OF  TABLES. 


147 


TABLE  Y 1 1 . — Continued. 


"Weight  of  Round  Iron  in  Pounds  and  Ounces. 


r 

11  feet.  | 

11  feet. 

12  feet. 

13  feet. 

14  feet. 

15  feet. 

1«  feet. 

17  feet. 

it  fen. 

1 Ditmeter 
m lzcnes. 

lbs. 

o*.  i 

lbs. 

OS. 

lbs. 

OS. 

lbs. 

OS. 

lbs. 

oz. 

lbs. 

oz. 

lbs. 

oz. 

lbs. 

OS. 

lbs. 

oz. 

|'  * 

6 : 

11 

7: 

5 

8: 

Q 

8 : 

11 

9 : 

5 

10  : 

0 

10: 

11 

11  : 

: 5 j 

11  : 

: 14 

* 

10: 

7 

11  : 

8 

12: 

8 

13: 

9 

14  : 

10 

15  : 

10 

16  : 

11 

17  ; 

; 12 

IS: 

: 13  | 

L *_ 

15  . 

0 

16  : 

8 

18: 

0 

19  : 

8 

21  : 

0 

22: 

8 

24  : 

; 0 

25  ; 

: 8 

■ 27 : 

: C 

r * 

20: 

6 

22  : 

7 

24: 

7 

26  : 

7 

28  : 

8 

30  : 

8 

32  : 

: 8 

34  ; 

: 9 

36: 

: 10  \ 

X 

25  : 

8 

29  ; 

3 

31 

13 

34  : 

8 

37  : 

3 

39  : 

; 13 

42  : 

: 8 

45  ; 

; 2 

47  ; 

: 12 

S’ ]* 

33  : 

9 

37  : 

0 

40:  5 

43  : 

11 

47  : 

0 

50  : 

; 6 

53  ; 

; 12 

57  ; 

: 2 

60  : 

: 8 1 

41  : 

11 

45  : 

14 

50 

1 

54  : 

4 

58 : 

6 

62  : 

: 9 

66  : 

; 12 

70 

: 14 

75 

: 1 

1 ^ 

50 : 

3 

55  : 

4 

60 

4 

65: 

4 

70: 

6 

75  : 

; 6 

80  : 

: 5 

85 

: 5 

90 

: 4 j 

1 l*. 

59  : 

11 

65  : 

11 

71 

10 

77  • 

10 

83: 

9 

89  ; 

: 9 

95  ; 

; 10 

101  : 

: 8 

107 

= 8 1 

1 

70 

2 

77  : 

2 

84: 

2 

91  : 

2 

98: 

3 

105  ; 

; 3 

112  : 

; 3 

119  : 

: 3 

126 

: 3 j 

j IX 

81 

5 

89 

7 

97 

9 

105  : 

11 

113  : 

: 13 

121  ; 

; 15 

130  : 

: 0 

138 

: 2 

146 

: 4 | 

ix 

93: 

5 

102 

11 

112 

0 

121  : 

5 

130  ; 

11 

140  ; 

; 0 

149  ; 

: 5 

158 

: 11 

168 

: 0 j 

2 

106 

2 

116 

12 

127 

6 

138  : 

0 

148  : 

: 10 

159  : 

L 4- 

169  : 

: 14 

180 

: 8 

192 

: 2I 

2X 

120 : 

132: 

144: 

156: 

168  : 

180  : 

192; 

: 0 

204 

: 0 

216 

‘0| 

2X 

134  : 

7 

147: 

13 

161  ; 

: 5 

174  : 

; 11 

188  : 

: 2 

211 

: 10 

215  ; 

: 0 

228 

: 8 

242 

: 0| 

2X 

149  : 

12 

164  ; 

; 12 

179  ; 

: 12 

194  ; 

: 11 

209  ; 

; 11 

224 

: 11 

239 

: 10 

254 

: 10 

269 

:,nj 

2 % 

166  : 

14 

183  : 

9 

200  ; 

; 4 

216  : 

; 15 

233  : 

: 10 

250 

: 5 

267 

: 0 

283 

: 10 

300 

: 5 

2X 

187  : 

8 

206  : 

; 4 

225; 

: 0 

243  : 

; 12 

262 

: 8 

281 

: 4 

299 

: 11 

318 

: 12 

337 

: 8 

2X 

200  : 

: 12 

220: 

13 

240  ; 

: 15 

261  ; 

; 1 

281  ; 

: 1 

301 

: 2 

321 

: 3 

341 

: 5 

361 

: 6 

2X 

219  : 

6 

241  : 

5 

263  ; 

: 5 

285  : 

: 4 

307  ; 

: 2 

329 

: 2 

351 

: 2 

373 

: 0 

395 

: 0 

3 

238  : 

; 14 

262  ; 

; 12 

286 

: 10 

310  ; 

: 8 

334; 

: 6 

358 

: 4 

383 

: 3 

406 

: 1 

430 

: 0 

3X 

259  : 

; 5 

2S5  ; 

; 4 

311 

: 3 

337  : 

: 1 

363: 

: 0 

388 

: 14 

414 

: 12 

440 

: 11 

466 

: 10 

3X 

280  : 

: 7 

308  ; 

; 7 

336 

: 8 

364; 

: 8 

392 

: 9 

320 

: 10 

448 

: 10 

476 

: 11 

504 

: 11 

3X 

302; 

; 7 

332; 

; 10 

362 

: 14 

392 

: 2 

423: 

: 6 

453 

: 10 

483 

: 13 

514 

: 1 

544 

: 4 

3X 

325  ; 

; 2 

357  ; 

; 10 

390 

: 2 

i 422 

: 10 

455 

: 3 

487 

: 10 

520 

: 3 

552 

: 12 

585 

: 5 

3X 

348; 

: 14 

383 

: 12 

418 

: 10 

453 

: 8 

488 

: 6 

523 

: 5 

558 

:*  4 

593 

: 2 

628 

: 0 

3X 

373  ; 

: 5 

410 

: 11 

448 

: 0 

4S3  ; 

: 5 

| 522 

: 11 

560 

: 0 

597 

: 5 

634 

: 11 

672 

: 0 

3X 

398; 

: 10 

438: 

: 8 

478 

. 6 

518 

: 4 

558 

: 2 

598 

: 0 

637 

: 13 

677 

: 11 

717 

: ft 

4 

1 424; 

; 10 

467  : 

: 2 

509 

: 9 

552 

: 1 

594 

: 8 

637 

: 0 

676 

: 6 

621 

: 14 

“64 

: 6 

4* 

j 451 

: 11 

496 

: 14 

542 

: 2 

587 

: 5 

632 

: 7 

677 

: 10 

722 

: 12 

761 

: 0 

813 

: 2 

4X 

( 479 

: 8 

527 

: 7 

575 

: 6 

523 

: 6 

671 

: 5 

719 

: 4 

767 

: 3 

815 

: 2 

863 

: 2 

508 

: 3 

559 

: 0 

j 609 

: 13 

660 

: 10 

| 711 

: 6 

762 

: 4 

813 

: 0 

863 

: 15 

914 

: il 

4X 

537 

: 11 

591 

: 6 

l 645 

: 2 

698 

: 15 

7V2 

: 10 

806 

: 6 

860 

: 3 

913 

: 14 

| 967 

: 11 

4^ 

567 

: 15 

624 

: 11 

| 681 

: 8 

738 

: 3 

795 

: 0 

851 

: 13 

908 

: 10 

965 

: 6 

! 1022 

: 3 

4X 

599 

: 0 

1 658 

: 14 

718 

: 12 

| 778 

• 11 

S3  ft 

• 10 

i 898 

: 8 

958 

: 6 

1018 

: 5 

1078 

: 3 

4X 

630 

: 15 

694 

: 0 

1 757 

l 

: 2 

820 

: 3 

883 

| 946 

: 6 

1009 

: 8 

1072 

: 10 

1 1135 

: 12  | 

6 

667 

: S 

1 734 

: 5 

1 801 

: 0 

l 867 

: 12 

i 934 

: 8 

1 1001 

: 5 

I 1068 

: 0 

1 1134 

: 12 

I 1201 

■ gj 

TABLE  VIII. 


This  table  exhibits  at  one  view  the  weight  and  strength  of  various 
kinds  of  timber,  as  ascertained  by  actual  experiments. 

The  estimates  of  the  weight  are  made  when  the  timber  is  well-sea- 
soned and  dry,  but  not  kiln-dried. 

There  may  appear  to  be  a discrepancy  between  the  strength  of  tim* 
ber  here  given,  and  that  found  in  the  concluding  remarks  on  Bridge 
Building ; but  it  must  be  borne  in  mind  that  the  calculations  there 
made  were  on  the  greatest  safe  strain  to  which  timber  should  be  sub 
mitted  for  a long  time  without  injury,  or  even  impairing  its  elasticity, 
while  the  figures  here  given  show  the  absolute  strength,  or  the  point 
of  breakage. 


TABLE  Y ill. 

Weight  and  Strength  of  Timber. 


Kind  of  Timber. 

Weight  com- 
pared with  wa- 
ter— water  be- 
ing 10»0. 

No.  of  lbs.  in 
a cubic  foot. 

No.  of  cubic 
feet  in  a toil. 

Greatest  ten- 
sile strength  of 
a square  inch 
in  lbs. 

i Greatest  safe  strain 
'upon  a beam  resting 
upon  the  ends,  and 
j loaded  in  the  middle  ; 
per  square  inch  in  lbs. 

White  Oak,  American 

672 

42 

53 

10200 

800 

'English  Oak  

930 

58 

38 

11800 

875 

Beech 

850 

42 

45 

12200 

1000 

Sycamore 

600 

38 

59 

9600 

720 

i 

Chestnut 

610 

38 

59 

10650 

■ 

650  j 

Ash 

845 

52 

43 

14100 

950 

Elm 

670 

42 

53 

9700 

700 

| Walnut 

670 

42 

53 

8S00 

675 

j Poplar 

380 

34 

66 

5900 

380 

Cedar  

660 

33 

68 

7400 

400 

White  Spruce 

1- 

550 

34 

66 

7200 

550 

1 

White  Pine 

590 

37 

60 

7300 

575 

I 

Yellow  Pine 

460 

28 

80 

11800 

770 

Pitch  1 i ue 

660 

H 

54 

9800 

750 

Fir i 650 

34 

66 

9500 

675 

Strength  of  Timber. 


This  is  determined  only  by  actual  experiment,  and  different  experi- 
ments give  different  results,  according  to  the  circumstances  under  which 

they  are  made.  Timber  is  much  stronger  length-ways,  with  the  grain, 
(148) 


EXPLANATION  OP  TABLES. 


149 


than  side-ways,  across  the  grain ; the  former  is  called  its  tensile  strength 
and  the  latter  its  lateral  strength . In  the  tables  on  page  148  we  give 
both  the  greatest  tensile  and  the  greatest  lateral  strengths  of  different 
kinds  of  timber : we  now  propose  to  give  some  rules  and  data,  by  .which 
any  intelligent  mechanic  can  make  his  own  calculations ; these  data  beifig 
based  upon  the  strength  of  good  white  oak,  to  which  all  other  kinds 
of  timber  may  be  compared.  In  the  table  it  is  stated  that  the  greatest 
safe  strain  of  oak  timber,  resting  upon  both  ends  and  loaded  in  the 
middle,  is  800  pounds  to  the  square  inch : that  is,  a stick  1 foot  long 
and  1 inch  square  will  support  800  pounds  in  the  middle ; and  while 
this  estimate  is  correct,  being  based  upon  actual  experiment,  yet  it  ap- 
proaches very  near  the  breaking  point,  and  is,  no  doubt,  too  high  for 
practical  estimates.  We,  therefore,  propose  600  pounds  to  the  square 
inch,  as  safe  and  practical,  and  will  proceed  to  give  our  rules  accordingly. 

When  one  end  of  a beam  is  fixed  permanently  in  a wall  and  loaded 
at  the  other  end,  if  the  beam  be  made  twice  as  long  it  will  bear  half  the 
load,  and  if  the  same  beam  be  supported  at  both  ends  it  will  bear  eight 
times  the  load.  If  a beam  be  supported  at  both  ends  and  loaded  in  the 
middle,  it  will  bear  twice  the  load  when  evenly  distributed  throughout 
its  entire  length.  We  are  now  to  consider  the  comparative  strength 
of  different  sticks  of  timber,  when  supported  at  both  ends  and  loaded 
in  the  middle,  when  varying  in  length,  width,  and  thickness. 

Rule.- — Multiply  the  square  of  the  perpendicular  by  the  horizontal, 
in  inches,  at  the  center,  and  that  product  by  600 ; then  divide  by  the 
length,  in  feet,  and  the  quotient  will  be  the  load,  in  pounds,  at  the 
middle  of  the  timber. 

Wote.—This  rule  may  require  two  corrections,  although  it  is  often 
used  without  them : 

1.  Deduct  the  weight  of  the  timber,  which  is  estimated  at  50  pounds 
to  the  solid  foot. 

2.  When  the  timber  is  more  than  20  feet  long,  multiply  by  50  lest 
for  every  5 feet : that  is,  from  20  to  25  by  550 ; from  25  to  30  by  500 ; 
from  30  to  35  by  450 ; from  35  to  40  by  400,  and  so  on. 

See  examples  on  next  page. 


150 


CARPENTRY  MADE  EASY. 


Examples . 

1.  What  is  the  strength  of  an  oak  joist,  3 inches  thick,  12  inches 
wide,  and  16  feet  long,  when  set  on  its  edge? 

12  x 12  x 3 = 432 
600 

16)259200 

16200 

200  Weight  of  timber. 

16000  lbs.  Am. 

2.  What  is  the  strength  of  the  same  joist,  when  lying  on  its  side? 

3 X 3x12  = 108 

600 

16)64800 

4050 

200  Weight  of  timber. 

3850  lbs.  Am. 

3.  What  is  the  strength  of  an  oak  joist,  8 inches  wide,  2 inches  thick, 
and  16  feet  long,  when  standing  on  its  edge? 

8x8x2  = 128 

600 

16)76800 

4800 

88  Weight  of  timber. 

4712  lbs.  Am. 

4.  What  is  the  strength  of  a beam  12  by  12  inches,  and  20  feet  long  ’ 

12  x 12  x 12  = 1728 
600 

20)1036800 

61840 

1000  Weight  of  timber. 


50840  lbs.  Am. 


EXPLANATION  OP  TABLES. 


151 


5.  What  is  the  strength  of  an  oak  joist,  3 inches  thick,  12  inches 
wide,  and  24  feet  long,  standing  on  its  edge  ? 

i 

12  X 12  x 3 = 432 
550 

21600 

2160 

24)237600 

9900 

300  Weight  of  timber. 

9600  lbs.  Ana . 

6.  What  is  the  strength  of  an  oak  joist,  3 inches  thick,  12  inches 
wide,  and  32  feet  long,  standing  on  its  edge? 

12  x 12  x 3 ^ 432 

500 

32)216000(6750 

192  400  Wreight  of  timber. 

"~240  6350  lbs.  Ana. 

224 


160 

160 

In  all  the  above  examples  the  strength  of  the  timber  is  estimated  for 
a load  at  the  center,  where  the  strain  is  the  greatest.  If  the  load  is 
nearer  one  end,  the  same  stick  will  support  a greater  weight,  and  the 
strain  will  be  in  proportion  to  the  product  of  the  two  ends  of  the  stick, 
measured  from  the  point  where  the  load  rests. 

For  example,  if  the  stick  is  10  feet  long,  and  the  load  rests  at  the 
middle  point,  then  the  two  ends  are  5 feet  each,  and  the  strain  is  equal 
to  5 X 5 = 25 ; but  if  the  load  is  3 feet  from  the  center,  one  end  is  8 
feet  and  the  other  2 feet,  and  the  strain  is  equal  to  8x2  — 16,  and  we 
form  a proportion,  thus : 

25:16::  10 : 6.4 ; so  instead  of  dividing  by  10  we  divide  by  6.4,  and 
find  the  true  result,  and  so  on  in  similar  cases. 


152 


CAKPENTRr  MADE  EASY. 


Examples . 

1.  What  weight  will  an  oak  stick,  10  feet  long,  10  inches  wide,  and 
5 inches  thick,  sustain,  3 feet  from  its  center,  the  stick  resting  on  its 
edge? 

10  x 10  x 5 = 500 
60Q 

6.4)300000(46875 

256  175  Weight  of  timber. 

440  46700  lbs.  Am. 

384 


560 

512 


480 

448 

320 

320 


2.  What  load  will  an  oak  stick  of  timber  sustain,  10  inches  wide,  3 
inches  thick,  and  18  feet  long,  resting  on  its  edge,  the  load  being  placed 
4 feet  from  the  center  ? 


10  x 10  x 3 = 300 
600 

14.4)180000(12500 

144  175  Weight. 

360  12325  Ans. 

288 


720 

720 


81: 65:  :18  s 
65 

90 

108 

81)1170(144 

81 

360 

324 

36.0 


For  evidence  of  the  correctness  of  these  rules,  see  Eatorfs  High  School 
Arithmetic  and  ZelVs  Encyclopaedia , 2d  Vol.,  p.  316. 


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Forty-Four  Plates  and  200  Figures. 


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of  Ilip  Rafters.  Table  8: — Octagonal  Roofs.  Table  4: — Length  of  Braces.  Table 
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Round  Iron.  Table  8: — Weight  and  Strength  of  Timber. 

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PRICE,  $5.00  PER  COPY. 

ADDRESS,  W.  E.  BELL, 

OTTAWA. 

ILLINOIS, 

O 

Liberal  IDisco'a.rrt  to  Agents. 


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Millersburg,  Ohio,  June  1373. 

ADDRESS,  W.  E.  BELL, 

OTTAWA, 


ILLIITOIS. 


6 3L1  aiDttltf 
1 Edz 


27  8/ 


